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MATHEMATICAL QUESTIONS, 



WITH THEIE 



SOLUTIONS. 



FEOM THE "EDUCATIONAL TIMES.' 



WITH MANT 



IPsptrs wxH Sahitaxts xmt irublis^eb in t^t '* (EbucHtbnal Sinus/' 



EDITED BT 

, W. J. MILLEE, B.A., 

MATBRMATIGAL MA8TBB, HUDDXB8FIBLD COLLKOR. 



VOL. XI. 

FROM JANUARY TO JUNE, 1869. 



LONDON: 

0. p. HODGSON & SON, GOUGH SQUARE, 

FLEET STOEBT. 
1869. 



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*.* 



5it* Volumes L to X. may still he hady price 6s, 6d. each. 



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LIST OF CONTRIBUTOES. 



rris, Berks. 



s in the UniTenity of Cam- 
of France. 

r Office, Southampton. 

ibridge. 

d; raddent of the Queens- 



Gf^Ot Cambridge, 
riore di Milana 
ral Military Aoad., Woolwich. 



kjhool. 



Queen's University, Belfest. 
n the B. M. Acad., Woolwich. 



St. John's College, Sydney* 
. (Cadets') CoU., Sandhurst. 

iversity College, London. 

50. 

ge, Cambridge, 
r Warrington. 
Grammar School. 

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IV 



Lilosophy in the 

id Architectiire, 
>ridge. 

Dublin. 

; Mathematical 



BTAJTLEY, ABCHES. 

Symbs, B. W., BJL, London. 

Stlyesteb, J. J^ F.B.S^ Professor of Mathematics in the Boral Military Acaden^, 
Woolwich ; Corresponding Member of the Institute of France. 

Tait, p. G.^V.A., Professor of Natural Philosophy in the Uniyersity of Edinburgh. 

Tasleton, FBAKCis A., M.A., Pellow of Trinity College, Dublin. 

Tatlob, C, M JL, Pellow of St. John's College, Cambridge. 

Taxlob, H. M.. B.A., Pellow of Trinity College, Cambridge; Vice-Principal of the 
Boyal School of Naval Architecture, South Kensington. 

Taylob, J. H., BA., Cambridffe. 

Tbbay, Septimus, BJl., Heaa Master of Rivington Grammar SchooL 

Thomson, P. D., M JL, St. John's College, Cambridge. 

ToDHUNTEB, IsAAC, P.B.S., St. John's College, Cambridge. 

ToMLiNSON, H., Christ Church College, Oxford. 

ToBEij.1, Gabbiel, Naples. 

TOBBY, Bev. A. P., MJl.. St. John's College, Cambridge. 

TowNSEVD, Bev. B., MA., P.B.S., Pellow of Trinity Colleffe, Dublin. 

TncKEB, B., M.A., Mathematical Master in University College School, London. 

WaIiEEB, J. J., MJL, Mathematical Master in Univeraity College School, London. 

Walmsley, John, Woolwich Common. 

Wabbbn, B., ma.. Trinity College, Dublin. 

Watson, Stephen, Haydonbridge, Northumberland. 

Whitwobth, Bev. W. A., M.A., Professor of Mathematics and Natural Philosophy, 
Queen's College, LiverpooL 

Wilkinson. Bev. M. M. U., Beepham Bectory, Norwich. 

Wilkinson, T. T., P.B.A.S., Burnley. 

Wilson, J. M., MA., P.G.S., Pellow of St. John's College, Cambridge; Mathe- 
matical Master in Bugby School. 

Wilson, Bev. B., D.D., Chelsea. 

Wolstbnholme, Bev. J., MA., Pellow of Christ's College, Cambridge. 

WooLHOUBE, W. S. B., P.BA.8., &c., London. 



CbfUributors deceased since the PubUeation qf Vol. I, 
Db Mobgan, G. C. ma.: Holditch, Bev. H., MA.; Lea, W.; O'Callaohan, 
J. ; PuBKiB8,^H. J., BA. ; Pbouhbt, E. ; Sadlbb, G. T., P.BA.S.; Wbioht, 
Bev. B. H., M.A. 



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CONTENTS. 



iWatfttmatiral ^aptrsf, ^r* • 

No. Page 

70. To find the number of permntations of n things taken r to- 

gether. By C. R. RiPPiN, M.A 40 

71. Note.on Question 2740. By Professor Oailby 49 

72. To express the distance between the centres of the circmnsoribed 

and inscribed circles in terms of the radii of those circles. 
By J. Walmslet, B.A 63 

73. Note on the late Judge Hargreave's Solution of the Quintio. 

By the Rev. T. P. Xi&kkan, M.A., F.R.S 82 



1587. Apollo and the Muses accepted the challenge of Jove, to vary 
the arrangement of themselves on their filzed and burnished 
couches at his evening banquets, till eveiy three of them 
should have occupied, once and once only, every three of the 
couches, in every and any order. In how many days, azid 
how many ways, did they accomplish the feat, keeping one 
arrangement of themselves through all the solutions P Re- 
quired two or more of these solutions, clearly indicated, so 
as to save space, by cyclical operations 65 

1819. From two fixed points on a given conic pairs of tangents are 
drawn to a variable confocal conic, and with the fixed points 
as foci a conic is described passing through any one of the 
four points of intersection. Show that its tangent or normal 
at that point passes through a fixed point 83 

1878. If a line of given length be marked at random in n points, 
and broken up at those points, find the chance that the sum 
of the squares on the parts shall not exoeed one-nth of the 
square on the whole line 17 

1912. If a and b be the points of contact, with a curve of the third 
class, of a double tangent ; and if this tangent be intersected 
in m, n, p by the three tangents to the curve which can be 

drawn from any point in the plane, then , — '— — 1^ = — > 

hn,bn ,hp p^ 

where p^, p^ are the radii of curvature at the points a and 6 41 



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VI CONTENTS. 

No. Page 

2101. Trouver leB conditions n^oeesaireB et soffiBantes pour que lea 
qoatre raoinea d'one ^nation dn qnotri^me degr6 forment un 
qnadrilat^re inscriptible. Trouver la aarface et le rayon de 
oe qnadrilai^re 46 

2108. Required analogaea in Solid Gteometry to the following pro- 
positions in Plane Geometry : — 

(a,") The perpendionlara of a triangle meet in a point. 

(b.) The middle points of the diagonals of a quadrilateral 
are in one straight line. 

(c.) The circles whose diameters are the diagonals of a 
quadrilateral have a common radical axis. 

(d4 Every rectangular hyperbola circumscribing a triangle 
passes through the intersection of perpendiculars. 

(e.) Every rectangular hyperbola to which a triangle is 
self-conjugate passes through the centres of the four touching 
circles. 

(f.^ sin(A + B) BsinAcosB + cosAsinB. 

(gr.) The sum of the angles of a triangle «s two right angles. 

(h.) In any triangle !mA ^ rinB ^ mnO ^^ 

a c 

2109. Two lines are drawn at random across a convex dosed curve ; 
determine the chance of their intersecting 94 

2221. 1. Show how to determine the locus of the feet of perpen- 
diculars from a fixed point on the generating lines of a system 
of confooal quadrics. 

2. Prove that this locus is described by foci of the plane 
sections passing through the fixed point 64 

2244. Form 11 symbols into sets, 6 symbols in a set, so that every 

combination of 4 symbols shall appear once in the sets 97 

2801. A circle is drawn so that its radical axis with respect to the 
focus S of a parabola is a tangent to the parabola ; if a tan- 
gent to the circle cut the parabola in A, B, and if SG, bisect- 
ing the angle ASB, cut A3 in C, the locus of C is a straight 
line 31 

2345. Determine in trilinear coordinates the foot of the perpen- 
dicular from the point (a/, y', Z) on the line Ix + my + nz ^^0; 
also the reflexion of the same point with regard to the same 
line 30 

2369. Given a cubic curve K, and a point on it p; through P is 
drawn any transversal meeting E again in m and n, and on 
it is taken a point x such that the anharmonic ratio (jpTnvn) 
shall be equal to a given quantity. Prove that the locus of 
is a quartic curve with a point of osculation at p, touching 
K at j7 (counting for four points of intersection) and in four 
other points 43 

2456. If a, b, c be the three sides of a spherical triangle, and k the 

radius of its polar circle, prove the formula 
^^j , _^ (cos h cos c sec a— 1) (cos c cog a sec b— 1) (cos a cos h sec c— 1) ^^ 
4 sin « sin {s—a) sin (s— b) sin («— c) 



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CONTENTS. Vll 

No: Page 

2456. Show that i£ an ellipse pass through the centre of a hyper- 
bola, and have its foci on the hyperbola's asymptotes j (a) the 
hyperbola passes through the centre of the ellipse, (b) the 
axes of each curve are respectively tangent and normal to 
«ach other, and (c) the two axes which are also normals are 
equal to each other 31 

2486. Pind a point on an ellipse such that the normal produced to 

meet tbe curve again may cut off a maximum area ^. 91 

2494. On each side of a hexagon as base, a triangle is described by 
producing the two adjacent sides to meet, and a second hexa- 
gon is formed by joining the vertices of these triangles in 
order. Show that if either of these two hexagons can be in- 
scribed or circumscribed to a conic, the other can be circum- 
scribed or inscribed to a conic 44 

2513. Let A, B, 0, D, B, F be six points. Let BF, CD meet in a ; 
AB, CB in b; AB, CD in c; AF, CB in d; BF, DE in e; and 
AF, DB in /. Show that, if either of the hexagons ABCDBF, 
dbcd^ can be inscribed or circumscribed to a conic, the other 
can be circumscribed or inscribed to a conic 44 

2525. Three equidistant lines are drawn parallel to an asymptote of 
a hyperbola, and a triangle inscribed where the lines meet 
the curve. Then any line parallel to the other asymptote 
will be divided harmonic&lly by the three sides of the triangle 
and the curve t 20 

.2527. Find the locus of the intersection of two tangents to a circle, 
the chord of contact of which subtends a right angle at a 
fixed point ; and also of the middle points of a system of 
chordB of a circle, which subtend a right angle at a fixed 
point; and thence show that the envelope of a system of 
chords of a conic, which subtend a right angle at the focus, 
is another conic of coincident focus ; and that the locus of 
the middle points of a system of chords of a conic which sub- 
tend a right angle at a fixed point is another conic 93 

2533. Find the envelope of a line upon which two given conies inter- 
cept segments which have a common middle point ; and find 
also the locus of this middle point 84 

2539. If three conies cut one another so that every two of them 
have a common chord which is also a chord of a fourth conic, 
the other three chords of intersection of the three conies meet 
inapoint .■ 44 

2544. There are two values of d for which o, jS, 7, the roots of the 
cubic agfi + Sha^ -k- 3cx + d a fulfil the linear relation 
Aa-l- B/3+ C7 B 0. Show how to find the quadratic equation 
of which these values of d are the roots ^ 64 

2597. A right cone, whose weight may be neglected, is suspended 
from a point in its rim ; it contains as much fiuid as it can : 
show that the whole pressure upon its surface is 
, ,. sinacosd (cos(0-i-a)f$ 



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viii CONTENTS. 

No. Ft«e 

where h and 2a are the height and vertical angle of the cone, 
and 9 is determined from 3 sin 20 » 4 sin 2(0— a) 101 

259S. A nniform rod (mass «* fi) is placed inside a spherical shell 
(ma8s» m, radios ^ a), which, under the influence of gravity, 
rolls down the surface of a rough sphere (radius » r) ; find 
equations for the movements of the rod and shell, there being 
no firic^ion between them ; and hence show that, if a particle 
/K, is placed near the lowest point of a spherical shell (m) 
which performs small oscillations on a rough horizontal plane, 

the length of the simple equivalent pendulum is a 62 

2m + Ai 

In a plane, (1) on every line, there are two points harmonic 
conjugates to the conies through four points A, 6, G, D. They 
are real, if the product of the four triangles ABC, BCD, CDA, 
DAB, and the perpendiculars firom the points A, B, C, D on the 
line, is positive. Tf a line does not cut a conic, why are the 
conjugates on the line real, to any set of four points on the 
conic ? Give a rule for determining the nature of the points 
on a line, when the product of the triangles is (1) positive, 
(2) negative. 

(2.) Through any point, there are two lines conjugate to 
the conies inscribed in four lines. They are real in those 
regions of the plane formed by the four lines in which the 
product of perpendiculars from the point has the same sign 
as that of points in the conVex quadrilateral. Hence every 
point on or within a conic inscribed in the four lines lies in 
these regions 106 

2606. The developable circumscribing two surfaces of the second 
degree touches either of them along a curve, which is its 
intersection with another surface of the second degree 68 

2616w Let two intersecting tetrahedra have all their edges bisected 
by the same system of Ciurtesian axes, each axis through two 
opposite edges of each tetrahedron ; then the solid about the 
origin has the origin for its centre of figure 47 

2618. A pack of n different cards is laid, face downwards, on a 
table. A person names a certain card, that and all the cards 
above it are shown to him and removed ; he names another, 
and the process is repeated. Prove that the chance of his 
naming the top card during the operation is 

i_l + l_J.+ -.(nil" 70 

2622. The equation connecting the distances (ri, r2, rj) of any point 
on a Cartesian oval from the foci is 

(/3-7)o*ri+ (7-a) /bVj+ (a-0) 7^3 = 0, 
where a, /3, 7 are the distances of the foci from the triple focus 66 

2623. If a conic pass through the four points of contact of tangents 
to a cubic from a point (A) on the curve, and through two 
other points (B, C) on the cubic ; then A is the pole of BC 
with regard to the conic 68 



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CONTENTS. IX 

No. Page 

2636. 1. If A, B be two confooal oonics, and A! be a third oonio 
having doable contact with A, show that there exists a fonrth 
conic B' having double contact with B and confocal with A'. 

2. If A, B be two confocal quadrics, and A' be a third 
qnadric having continnous contact with A, show that there 
exists a fourth qnadric B' having continnons contact with B 
and confocal wi^ A! ^ 43 

2640. I. If a conic of given eccentricity e cironmBcribe a triangle 
ABC, the locns of its centre is 

{yz sin» A + gg sin^ B + a?y sin^ C)^ ^ (fr^. 
xyz (w + y + z) sin^ A sin* B sin* C 1— e* ' 
and (2) if it be inscribed in the triangle, the locns of the 
centre is 
(jF* sin* A+ ...+2pz sin B sin C cos A-f ...)* _ 4(e*- 2)* . 
xyz (x + y + z) sin* A sin* B sin* "" 1— e* 

the triangle of reference having its angular points at the 
middle points of the sides of the triangle ABC. 

3. Hence show that the locus of points with which as centre 
two similar conies can be drawn, one inscribed and the other 
circumscribed to the triangle, is two circles, one the circum- 
scribed circle of the triangle, and the other that which haa 
the centre of gravity and the centre of perpendiculars of the 
triangle as the ends of a diameter 25 

2642. If three surfaces of the second degree meet in four points, at 
each of which their three curves of intersection have a com- 
mon tangent, these four points lie in one plane 77 

2651. Determine the form of the solution of the differential equation 

^•^^'^S * '•'"^'^^ "" {•^'<'>±"'} ^ = <* ^^ 

2653. 1. If the centre of one of the four circles which touch the 
sides of a triangle self-conjugate with respect to a parabola 
lies on the directrix; show that its circumference passes 
through the focus. 

2. & the point of intersection of the three perpendiculars 
of a triangle inscribed in a parabola lies on the directrix $ 
show that the self -conjugate circle of this triangle paases 
through the focus of the curve 32 

2654. Determine the polar reciprocal of the quartic curve 

y«a;* + »*«*-H«^* + 2ay«(l» + wiy-»-w») «0 60 

2670. An observer, seated in the aisle of a cathedral and looking 
westward, sees the horizontal lines above the arches con- 
verging to a point before him, and consequently cutting the 
horizon at a certain angle. On his facing northwards, he 
sees that the same lines are now parallel to the horizon. 
Bequired the curve that they will appear to lie in, as he 
gradually turns his head through 90° 57 

2688. A variable triangle circumscribes an equilateral hyperbola, 
and is such that its nvne-^pomt circle passes through the 
centre of this curve ; prove that the locus of the centre of its 
cirCMwwonlmi^ circle is the hyperbola in question «., 59 

b 

7-' 



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:X CONTENTS. 

No. Fkffe 

2701. In a right conoid whose axis is the axis of «, show that the 
radii of principal cnrvatnre at the point (r cos 9, r sin 9, «) are 
giren hj the equation 

v(S)V-»S{"*(S)T-!'"-(S)T '" 

2712. Form 15 symbols into sets, 5 symbols in each set, so that 

every combination of 4 symbols shall appear once in the sets 97 

2718. Find in plomo the loons of a point F, snoh that firom it two 
given points A, C, and two given points B, D, subtend eqnal 
angles 83 

2719. 1. The six points which, in conjunction with any common 
transversal, divide harmonically the six sides of any tetra- 
stigm, lie all on one conic, which also paases through the 
three points of intersection of opposite sides of the tetrastigm. 

2. Of the six points, if those pairs which lie on opposite 
sides of the tetrastigm be joined by straight Unes, the three 
straightlines thus drawn are concurrent, and their point of inter- 
section is the pole of the transversal with respect to the conic 21 

2724. Three smooth rings, F, Q, B, are thread on an endless string 
of given leng^, and constrained to move on three straight 
rods, OF, OQ, OB. To investigate the motion oorrespondmg 
to a slight arbitrary disturbance of the system 75 

2726^ If a conic touch the sides of a triangle and pass through the 
centre of the circumscribed circle, the director circle of the 
conic will touch the circumscribed circle of the triangle 60 

2727. Two lines (a) and (h) in space touch a fixed quadric U, show 
that a variable line intersecting the two former and touching 
the latter generates another quadric 20 

2728. Given three planes and their poles with regard to a system of 
quadrics, the locus of centre js a right line 101 

2736. Find the curve whose radius of curvature and radius vector 
at each point are equal to one another, and prove that it is 
the second involute to a circle having an apse midway be- 
tween the cusp of the first involute (from which it is derived) 
and the centime of the circle 18 

2740. Prove that the envelope of a line given by an equation of the 
form L cos 2^ -i- M sin 2^ + P cos ^-t-Qsin^-t-B » 
may be obtained as the discriminant of a cubic equation, and 
form this equation 42 

2741. A moveable event (depending on the moon) happened in 
the year y, show that it wiU happen again in the year 
y + 19 (28t— 5a + 86) ; where amay be 0, 1, 2, 3 j h the num- 
ber of completed centuries since the year y which are not 
leap years; and t any arbitrary integer 52 

2745. Show that the average area of all the ellipses that can be in- 
scribed symmetrically in a portion of a parabola cut off by a 
double ordinate perpendicular to the axis, in parts of the area 
of the parabola, is -^w if (1) the centres be equably distri- 
buted over all possible positions on the axis, and ^ir if (2) 



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CONTENTS. XI 

No. Fige 

the intersections of the ohords of contact with the aoda be 
equably distribnted on the axis 24 

2756. Form 16 symbolB into sets, 4 symbols in each set, so that 
eveiy triad in the symbols shall appear once in the sets 97 

2757. If ^^j.^.^i 



». 9, 1 



(1). 

(2). 



show that each of the equations 

a« (ar-a?o)» + b« (y~yo)» ^ a^ (^-0^0'+ b» (y-yi)» 
(a?a?o+yyo-l)' " {xx^-i- ^f^-iy 
o^(x~aro)8 + ba(y~yo)8 ^ aH^-arja + b' (y^yQ' 

(a3fo-a'cy)«-(af-aro)'-(y-yo)* * (ayi-ariy)*- (af-a?i)»-(y-yi)«*" 

represents the right line L*-0 and a cubic curve 83 

2761. Find the conic envelope of the radical axis of two circles, one 
of which is fixed, whilst the other passes through a fixed 
point. If the radii of the circles are equal, find the foci and 
axes of the envelope 74 

2762. Find the envelope of an ellipse which has one vertex in the 
curve of a given parabola, and touches at its adjacent vertex 
the extremity of a fixed double-ordinate of the parabola 58 

2763. Give a geometrical interpretation for the relation between 
the invariants of two conies, viz., ee' » AA' 94 

2764. It is weU known that if A, P, P' be fixed points on a line (that 
is, if the distances AP, PP' are constant), and if PP' describe 
given fixed lines, say the directrices, then the point A de- 
scribes an ellipse having for its centre the intersection of the 
two directrices ; and the question hence arises : Given three 
positions of the point A and of the generating line APP*, to 
find the directrices ; or, what is the same thing. Given the 
lines 1, 2, 3, passing through the points A, B, G respectively, 
it is required to draw & line cutting the three lines in the 
points P, Q, B respectively, such that AP » BQ = OK 19 

2766. Three points move from a fixed point in three different but 
given directions, with given uniform but different velocities. 
Find the locus of the centre of the circle passing through them 22 

2769. From a point, taken at random within a triangle, perpen- 
diculars are drawn on the sides ; find the probability that the 
triangle formed by joining the feet of these perpendiculars is 
acute-angled 94 

2770. If a conic be drawn through S, S', the foci (both real or both 
impossible) of a given conic, and osculate the conic at a point 
P, the tangents at S, S' will intersect in the centre of curva- 
ture at P : 60 

2771. Prove that (1) if a straight line of length a be divided at 
random in two points, the mean value of the sum of the 
squares on the three parts is \o? ; (2) if the line be divided 
at random into two parts, and the longer part again divided 
at random into two parts, the mean value of the sum of the 



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Xll CONTENTS. 

No. Ffege 

squares on the three parte is ffo^ ; and (3) if it is an even 
chance that n times the sum of the squares on the parts in 
(1) is less than the square on the whole line, then 

»-i7TT73 ^« 

2773. When a given angle rolls upon a fixed parabola, the locus of 
its vertex is well known to be a hyperbola having the same 
focus and directrix as the parabola. Conversely, when a 
given parabola rolls within a fixed angle, show that its focus 
and vertex describe lines of the fourth and sixth orders re- 
spectively; find the actual equations of both, these curves, and 
thence show that the vertex of the fixed angle is a conjugate 
point on the former curve, and that the equation of the latter, 
when the given angle is a right angle, becomes 

a^ (a^ + y' + 3a?) — a« or x^y^ (a?* + y*) « a?, 

the sides of the given angle being the axes, and a equal to 
the distance of the focus from the vertex 9$ 

2774. A central conic {h^a^+ahf^ «= aPh^) is turned in its own plane 
about its centre through a right angle ; prove that the locus 
of the intersection of the normal at any point on the given 
conic with the same line in its new position is one of the two 
sextic curves 

(a^ + t/^^{a''(ai±yy±h\xTyy] « (a«+b3)'(«»-y')* 86 

2776. Through A, the double point of a circular cubic, draw AB 
perpendicular to the asymptote ; if chords be drawn to the 
curve subtending a right angle at the double point, show 
that there is a fixed point in AB at which also they subtend 
a right angle 63 

2778. Suppose fjL any quantity fractional or integral, and write 

C-«,;C-^ c^^ttSapa^ ^. MO^^^iy^X .- 

also let 71 be a positive integer such that /i and /jL + n hitve not 
different signs ; then it has been proved elsewhere, and may 
be assumed to be true, that the number of changes of sign in 
the series 

a^, fltl' — OoOj, Og^— OiOj, 

is em inferior Umdt to the number of imaginary roots in the 
equation 

c^* + cia?""* + <j2a?*"*+ — 0. 

1. Bequired to prove that the above theorem remains true 
when we write 

<,-a,. c^-^o,, c,-f^P^<H. c,-dt±Mt±^a, , 

y y{y-i-l) y{v+l)(y + 2) 

provided y is a positive integer, and ^— v+1, fjL + y do not 
differ in sign. 

2. Ascertain whether y being an integer is essential to the 
truth of the extended theorem 38 

2779.- Show that the famous equation between wrc, angle of cou' 



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CONTENTS. Xin 

No. Page 

Ungence, and perpencUciua/r on tcmgent to a corye, viz., 

I> + j~ » -J-* may be interpreted aa affirming the self -evident 

proposition that the radius of curvature is the distance between 
the tangent at any point of a cuire and the tangent at the cor- 
responding point of its second evolute ; and account for the 

connecting sign between p and -^ in the above equation 
hemg phis Q3id not nvinus 21 

2780. The envelope of the diord common to an ellipse and its circle 
of curvature is a curve of the fourth class which has three 
double tangents, one at infinity and the two others coincident 
with the conjugate diameters equally inclined to the axis. 
Prove this, examine the curve, and consider the cases of the 
hyperbola and parabola 62 

2781. Prove that the maximum value of the common chord of an 
ellipse and its circle of curvature, a and h being the semiaxes 
of the ellipse, is 

j-^^±_^{2(a«-a»6»+M)i-(a» + l.»)(2o?-b»)(ab»-a?)}* 28 

2782. 1. A given point is known to be within a certain circle of 
given radius, but unknown position; find the chance that 
another given point is also within the circle. 

2. Three given points are known to be within a certain 
circle, which is otherwise altogether unknown; determine 
the most probable position of its centre. 

3. Two given points are known to be within a circle, and a 
third given point is known to be outside it ; determine the 
most probable position of its centre 107 

2787. The value ofZ/T x^dx dAj da taken over the whole volume of 
a tetrahedron through one of whose comers the plane of (yz) 
passes, is ^V (oj^+bi^ + Ci^), where V is the volume, and 
Oi, &i, Ci are the distances of the middle points of the opposite 

I from the plane of reference 110 



2789. If two circles pass through the foci of an ellipse and touch the 
same variable tangent to the curve, the angle at which they 

intersect is constant and equal to 2 tan~^ - 62 

c 

2795. Find the average jK^uare of the distance between the centres 
of the inscribed and circumscribed circles of a triangle in- 
scribed in a given circle 64 

2797. In any complete quadrilateral, the radical axis of the three 
circles whose diameters are the diagonals will pass through 
the centres of perpendiculars of the four triangles formed by 
the four straight lines 65 

2798. To find the condition that a given point (x, y, z) may be ex- 
terior to the given conic 

S = (A, B, 0, F, G, H][ar, y, «)« - 41 



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XIV CONTENTS. 

No. Page 

2799. If n dioe are tlurown, what is the chanoe of an odd number of 
aoes tnmuig up P 61 

2800. Three oiroles paasing through a point P form a oironlar tri- 
angle ABC, and each side of this oironlar triangle or ifcs 
continuation is cut orthogonallj at the point P' by a cirole 
passing through P; prove that the three circles described 
about the triangles PAF, PBP", POP"' are coaxal 69 

2804. A straight pole stands verticaJly on a slope inclined to the 
south. If it be broken at random by the wind blowing in 
a given direction, so that the upper end of the pole rests upon 
the slope, determine the probable area of the triangle thus 
formed j and deduce the result for a horizontal plane 69 

2808. Four fixed tangents are drawn to a conic S; three other 
conies are drawn osculating S in any the same point P, and 
each passing through the ends of a diagonal of the circum- 
scribed quadrilatersJ : prove that the tangents to these conies 
at the ends of the diagonal meet in one point P', and that 
the locus of P' is a curve of the sixth degree and fourth class, 
having two cusps on every diagonal, and touching S at the 
points of contact of the four tangents 108 

2813. The equation of a conic referred to an axis and tangent at 
vertex being aa^ + hy^ + 2dsD » 0, if the conic be turned about 
the vertex in its own plane through a right angle, the locus 
of the intersection of any tangent in the original position 
with the same line in the new position of the conic is the 
(bicircular) quartic 

l){a(«2 + ya)+2d(a?+y)} (x^ + y^ ^^{x^y)\ 
and the corresponding locus for the normal is the sextic 

What does this latter equation become in the case of the 
parabola? 71 

2815. A cask contains a gallons of wine. Through a hole in the top 
water or wine can be let in at the rate of b gallons per mi- 
nute ; and through a pipe in the bottom, when open, the mix- 
ture can escape at the same rate. Suppose the discharge 
pipe is opened at the same instant that water is let in at the 
top, and t minutes afterwards the water is shut off and wine 
let in. Eequired the quantity of water in the cask at the end 
of ti minutes from the opening of the discharge pipe, and the 
length of time elapsed, both before and»afber wine was let in 
at the top, when the quantities of the two fluids in the cask 
were equal, supposing them to mingle perfectly 87 

2816. Find three square numbers in arithmetical progression, such 
that the square root of each (a) increased or else (js) di- 
minished by unity shall give three rational squares 88 

2818. Given a circle S and a straight line a not meeting S in real 
points ; O, O' are the two point-circles to which, and S, a is 
the radical axis ; two conies are drawn osculating S in the 
same point P, and having one focus at O, O' respectively : 
prove that the con esponduig directrices coincide 74 



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CONTENTS. XV 

No. Fftgo 

2821. Show that the mean ralne of the distanoe from one of the foci 
of all points within a given prolate spheroid is ^(3 + 6^), 2a 
being the axis and 6 the ecoentrioity 87 

2828. Show that on a chess-board the chance of a rook moving 
from one square to another without chang^g colour is f ; 
but that witliout altering the equality of the rmmher of the 
black and white squares, but only the manner of their distri- 
bution, the chance may be made equal to i 4B 

2828. A straight rod is divided into n parts in arithmetical progpres- 
sion, and equal particles are fixed at the points of division. 
If the system be made to vibrate about one extremity, deter- 
mine the average length of the pendulum, neglecting the 
weight of the rod 83 

2835. In a plane, if A, B are two points, and a point P describe a 
curve of the nth order, show that F', the intersection of the 
perpendiculars in the triangle PAB, will describe a curve of 
the 2nth order with three multiple points of the order n. 
Explain why the curve corresponding to a circle through the 
points A and B includes its reflexion to the line AB 86 

2839. In a triangle the bisector of the base is equal to the less side 
and also to one-half of the greater side j determine the three 
angles c 73 

2847. Prove that 

•»D-.,.*«-^ (l + 4,Wk)-j,-rTiXS^ where D - ± 67 

dx 

2870. Given three points A, B, G, and a conic j two points P, Q are 
taken on the conic such that the pencil A (BPQGJ is har- 
monic ; prove that the envelope of PQ is a conic touching 
AB, AO at points on the polar of A with respect to the given 
conic 83 

2875. Any tangent to a conic is, of course, divided in involution by 
three other tangents, and the lines joining their points of 
intersection to one of the foci of the conic ; prove that the 
distance between the double points of the involution subtends 
a right angle at the focus of the conic. The locus of these 
double points, the three tangents being fixed, is a cubic having 
a double point at the focus, whose nature I have not examined 106 

2877. If 

u^x + y + z + t, V *^fiD + gy + ha + kt, 

V wmygt + etw+txy + xyz, Y ^fyxt+gstx+htxy + kxyzi 
prove that the resultant of u, v, U, V is 

if-g) (f-h) (f-k) ig-h) (g-k) {h--h) 

xif+g-h-kyif+h-g-kyif+h^g-^h)^ 81 

2887. 1. If a point P on a conic be connected with any two fixed 
points A and B in its plane, all chords which are divided 
harmonically by PA and PB will be concurrent. 
2. The locus of the point of concurrence when P is variable 



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XVI CONTENTS. 

No. Page 

is another oonic which has double contact with the g^yen one 
onAB 100 

2893. If there be {n) quantities a, b, c, d, &'c., each of which takes 
independently a given nmnber of valaes aiagag. . . . b^bsbs. • . . 
&o, (the nmnber may be different for each), if we pat 

2 (a) ^a+h + c + d + Ac, 
and if for shortness we denote "the mean value ofx" by 
M (0), prove that 

M p (a)] = M (a) + M (b) +M (c) +&c. « 5[M (a)], 
M[2(a)]3« {2[M(a)]}»-.2[M((i)]«+2[M(a>)] 99 

2895. At any point on a cusped cubic a tangent is drawn meeting 
the cubic in a second point ; from the first point a Une is 
drawn touching the cubic in a third point. Prove that the 
lines drawn from the cusp to these points form with the 
cuspidal tangent a harmonic pencil 104 

2898. A circle rolls with internal contact on a fixed circle of half its 
radius ; prove that the envelope of any chord of the rolling 
circle is a circle which reduces to a point for a diameter 105 

2902. The circle passing through three points on a parabola, the 
normals at which co-intersect, always passes through the 
vertex; and if the point of co-intersection of the normals 
describe a coaxial conic having the vertex as centre and axes 
in the ratio of w : n, the locus of the centre of the circle will 
be a conic having the focus as centre, and axes in the ratio 
of 2w : n 104 



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MATHEMATICS 

VBOM 

THE BDtrOATIONAL TIMBS, 
WITH ADDITIONAL PAPBKB AND SOLUTIONS; 



1878. (PMposed by W. E. CuBVOVD, B.A.)— -If a line of given length 
be marked at random in n points, and broken np at those points, find the 
chance that the sum of the squares on the parts shall not exceed one-nth of 
the square aa the whole line. 

SohOion hy the Bev. J. Wolbtevholmb, MA.. 

Let «!, r» . • . . «,^ be the distances, in order of magnitude, of the n 
points of division from one extremity, and 1 the length of the line; then the 
limiting equation is 

«f + (««-*i)* + • . . . (l-^n)" < i 
or a?ft.... +a'i-*iar,-.... -.«^.ia?^+ 't±} < 0. 

Now I assume as obvious that a point (in space of n dimenaons) whose 
coordinates (rectangular or not) are Xi,Xf, ..•, x^, is a random point when 
4^1, jTj, .... dT^ are random lengths, and that the diance required is the value 
of ///'• • • • dxi dx2 .... dx^ when the limits are given by the equation 

a?! + ....+ar*-a?i»j-....+l!±l = (A), 

divided by the same integral when the limits are given by d?i < jr2 < d^ . . • • 
<x^<l, all positive. 

First, to find the quad-volume of the quadric (A). Its quasi-centre is 
given by 

Zxi^x^ « 2x^—Xi'-x^ » 2a?j— a?2~^4 =*•••• •= 2jf^— a?^_^— 1 » 0, 

1 2 
or flTs « 2*1, JTs B SoTi . . . . , or a?i « — -, a?j ^ .... 

YOL. XI. B 



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The equation of the surfiice referred to its eentre is then 



+ «n-«l*2-«-«'-'ii-l« 



2i»(ii-l-l) 

Now, transforming the axes (and to fiicilitate matters here I aasnme 
rectangular axes) so as to destroy the terms involving products, the resulting 
equation is 

«iXi + aaXa+ .... +"ji^« " ^JTJ^' 

where (see my Book of Mathematical PkoUems, Quest. 921) ai, o^, . . . . a„ 
are the roots of the equation 



3:-.2, 1, 0, 0, 

1, «-2, 1, 0» 0, .. 
0, 1, «-2, 1. 0, , 



0, 
0, 



0, 
0, 



0, 
0, 



.1. «-2, 1 
.0, 1. «-2 



= 0, 



the discriminant of 

(arj+ .... +a?")«-2(a?* + .... +dr"-a?iai-.. ..), 
and the product of these roots is » + 1. Also the value of fff,.,^y,„ rfX„^ 

where the limits are riven hy oiX?+ .. .. +o_ X„ ■» — - — -. is (Tod- 

* " fi(ft + l) 
hunter's Integral Calculus, Art. 274) 

1 f X )* * 1 ^ 1 _ S IT |*"__JL_ . 

(0,03 ...oj* «»(« + !)) r(irt+l) (»+l)M»(n + l)) r(i»+l)' 

and Jf^. . . . dxidx^ .... dir^, over all such positive values that arj < a?2 < «» 

• . . . < a?« < 1, is i or — i — ,. The chance required is then 
* [« r(n + l) 



4 5 IT l*» r(» + l) 

(»+i)* «»(»+!)) r(i»+i)' 



[The Proposer's solution is given on p. 83 of Vol. VI. of the Seprint.J 



2736. (Proposed hy Professor Stlvesteb.)— Find the curve whose 
radius of curvature and radius vector at each point are equal to one another, 
ahd prove that it is the second involute to a circle having an apse midway 
between the cusp of the first involute (from which it is derived) and the 
centre of the circle. 



Solution hy the Ret. James Wuite, M.A. 

The property of the second involute of a circle stated in this Question, viz., 
that when its apse is midway between the cusp of the first involute (from 



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19 



wbicb it 18 derived) and the centre of the circle^ its ndias vector and radius 
of carvatare will be equal, may be proved geometrically, as follows : — 

Let V be any p<nnt on this second involute whose 
apse is midway between O and B : it is required to 
prove that OV-PV. 

The second involute which commences at B, the 
cusp of the first, will cut PV in H, making HV»irr, 
(a being radius of the circle). Let QP=f • and draw 
OS perpendicular to VP; then OS » i, and SP * a. 

Now OV> = 0S« + VSa « ^+(VP-a)«, 



and 



raroBP + 4a =^ +?. 



VP = HP + ^ 




therefore 



OV = £1 + ? « VP ; 
2a * 2 



that is, radius vector (OV) » radius of curvature (VP). 

[Mr. Whitb states that the mam part of this proof is due to Mr. Cboftoh.] 

From this a simple method of finding a point on the second involute cor- 
responding to any one given on the first may be derived. 

Bisect the line joining the given point with the centre of the circle, and 
at the point of bisection erect a perpendicular. Also at the given point erect 
a perpendicular to the tangent from it to the circle ; the intersection of 
these perpendiculars is a point on the second involute which has its apse 
midway between the cusp of the first and the centre of the circle (the 
triangle OVP being isosceles). 

Take a point (H) on the perpendicular to the tangent distant from the 
one found one-half the length the radius of the circle, and it is a point on 
the second involute which commences at the cusp of the first. 

[Professor Sylyzsteb sends the following remarks on this question :— > 
The envelope of a set of half-pitch second involutes to any given circle is a 
drcle concentric with the first. Consequently anjf circle corresponds to a 
singular solution of the clifierential equation which defines the curve in 
question; and thus we obtain another and obviously true solution of the 
question.] 



2764. (Problem by A. Smith, Q.C.)— It is well known that if A, P, P' be 
fixed points on a line (that is, if the dirtances AP, PF are constant), and if 
P, F describe given fixed lines, say the directrices, then the point A describes 
an ellipse having for its centre the intersection of the two directrices ; and 
the question hence arises : Given three positions of the |X>int A and of the 
generating line APP', to find the directrices; or, what is the same thing. 
Given the lines 1, 2, S, passing through the points A, B, C respectively, it is 
required to draw a line cutting the three lines in the points P, Q, B respec- 
tively, such that APsBQaCB. 



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20 

SoliOkM hff Abohbk SsAiruET. 

Let p, qfT be any three correspondiDg points on the three given Hnes ; 
that is to say, let the three segments Ap, Bg, Cr (set off in assigned direc- 
tions) be equal to one another. Then the connector pq will, by a well4Knowa 
theorem, envelc^ a parabola of which Ap is a tangent ; in like manner, the 
connector pr will envdope another parabola having the tangent Ap in 
common with the first. The remaining two common tangents of these para- 
bolas will manifestly be the two required directrices. The problem is thus 
reduced to the well-known one (soluble by rule and compass; of finding the 
common tangents to two parabolas which touch a given line. 



2525. (Proposed by G. O. RAKLO]!r.)—>Three equidistant lines are drawn 
parallel to an asymptote of a hyperbola, and a triangle inscribed where the 
lines meet the curve. Then any line parallel to the other asymptote will be 
divided harmonically by the three sides of the triangle and the curve. 

Solution by the Rev. J. Wolstenholmb, M.A. ; the Psofobxb ; €md othen. 

First to find the locus of a point on any line in a given direction forming 
a harmonic range with the points where it meets the sides of the triangle of 
reference. 

If (/ + Ar) a? + (»i + ifc) y + (n + Ar) « = be any such straight line, then for 
the point (P) in question we have («i + Ar)y = (»+A?)a (if A'B'CP be 
harmonic), and the locus is 

. (Za? + «^ + n2)(y-«) + (» + y + 2:)(»«-«iy)=0 (1); 

or, in another form, 2(wi— »)y«— (»— Qza?— (?— «i)a?y = (2). 

Equation (2) shows that this is a conic circumscribing the triangle, and 
(1) that the asymptotes are parallel to lx-^m^-¥nz^Q and to y»«; the 
latter of which being the bisector of the side BC, shows that the three 
straight lines through A, B, parallel to one asymptote are equidistant. 
Hence the truth of the theorem. 



2727. (Proposed by the Rev. R. Townsend, P.R.S.)— Two lines (a) atid 
{h) in space touch a fixed quadric U, show that a variable line intersecting 
the two former and touching the latter generates another quadric. 



Solution hy W. S. Bubnsidb, M JL. 

It will be convenient to write the equation of the quadric IT under the 
form aa^ + hy^ + 2/nxy + 2trzw » 0, where the planes z and w plainly touch 
the surface, and the line (a?, ^) is their chord of contect. Whence we may 
consider {x, z) and (y, w) as the tangent lines (a) and (b). This being so, 
the condition that u variable line x =^\z, » » juy, should touch the quadric IT 
is ab\^ = (n\ + fwy, giving for the surfiuse traced out by it the equation 
abo^y^ s {^QDy -f rztaY, which resolves itself into the two constituents 
(» + ^ab) xy + rz» » 0, which represent two quadrics. 



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2779* (Proposed by Profeseor STi.yx8TB]|.)-— Sbow that the fkmotui 
equation between arc, angle of contingence, and perpendicular on tangent to 

a correi viz., p -¥ ^ »> — , may be interpreted as affirming the eelf-emdemt 

propoeiiion that the radios of car^atnre is the distance between the tangent 
at any point of a curve and the tangent at the corresponding pcnnt of its 

second evolute; and account for the connecting sign between p and -jB. in 
the above equation hemgplne and not minus. 



Solution ly J, J. Walkbb, M.A. 

1. Let (r, p, fp) refer to the given curve ; and let {ri, p^ ^i), (»Vt />& ^ 
refer similarly to its first and second evolutes respectively. Then it is evi- 
dent that p +|>s = pt 

where p — ^ is the radius of curvature of the given curve. 

Now "" ^ = g. ana P,-g-$ = ^ 
whence p+g_p»g. 

2. In what precedes, the case has been considered of p>p. But if p<p, 
t. e.> p—p% — pt BB p increases pi decreases, as is evident on drawing the 

figure; so that P2 — - 3^^ » — ^. Hence the sign between j? and J? 

is in both cases pUte. 



2719* (Proposed by Professor Whitwobth.)— 1. The six points which, 
in conjunction with any common transversal, divide harmonically the six 
sides of any tetrastigm, lie all on one eonic, which also passes through the 
three points of intersection of opposite sides of the tetrastigm. 

2. Of the six points, if those pairs which lie on opposite sides of the 
tetrastigm be joined by straight lines, the three straight lines thus drawn 
are concurrent, and their point of intersection is the pole of the transversal 
with respect to the conic. 

I. Solution by W.S.MoCay,B.JL 

It is not hard to prove this directly by Pascal's theorem ; but it is easier 
to consider the line at infinity, then we know (1) that the locus of centres 
of conies through the four points is a conic through the six middle points of 
the lines of the system, and that three other points on the locus are the 
three intersections of opposite lines of the system, and (2) that the centre of 
this conic is the intersection of the three Unes joining the middle points of 
opposite lines ; for these three lines meet and bislBct each other in the centre 
of gpravity <^ equal weights at the four vertices of the tetrastigm. 



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II. SohHon hy F. D. Thousov, M.A. 

1. This follows at once from tho well known theorem, that if a system of 
oonics he described about a given qaadrilateral, the locos of the pole of a 
fixed line with respect to these conies is another conic. 

This theorem may be proved analytically, as in Salmon ; or geometrically, 
as follows : 

The given transversal is cut in points in involution by any conic of the 
system, and by the opposite sides of uie quadrilateral ; and the double points 
of the involution are the points where two conies of the system touok the 
transversal. 

Now, for each of the points where the transversal cuts the required locus, 
the pole lies on its )poIar, and therefore must be the point of contact of the 
transversal with a conic of the system. 

Hence the only points in which the transversal meets the required locus 
are the two double points of the involution determined by the intersections 
of the transversal and the opposite sides and diagonals of the quadrangle. 

Hence the required locus is a conic section, and it is easily seen to pass 
through the points mentioned in the Question. 

2. The former part of this proposition may be derived by projection from 
well known properties of the middle points of the ades of a quadbUateral ; or 
directly, as follows : 

Let the transversal meet the sides in P, Q, B, 
S, T, V; and let the points of harmonic section 
be p, q, r, s, t, v. Then 

{B;?AP} = {D*AS}; 
therefore T{BpAP] « T{D»AS} 

-T{B*AP}; 

therefore Tps is a straight line. Similarly, Tqr, 
yrs, Yqpt B.8V, Qpv, &c. are straight lines. 
Hence, by the harmonic properties of a quadrilateral, pr and sq intersect in 
the pole of TV with respect to any conic through pqrs, and therefore the 
intersection of pr and sq is the pole of the given transversal with regard to 
the conic of the Question. Similarly, the interaection of sq and vt is the pole 
of PR, ». e. of the transversal, with respect to the same conic. Hence sq, pr, 
vt meet in the pole of the transversal. 




2766. (Proposed by S. Watson.) — Three points move from a fixed point 
in three different but given directions, with given uniform but different 
velocities. Find the locus of the centre of the circle passing through them. 

I. Quaternion Solution by W. H. Lavbbtt, B.A. 

Let a, jB, 7 be the unit-vectors in the directions of motion, and at the end 
of any time let ^^jo, kx^fi, kx^ be the vectors to the moving points.; so 
that, p being the vector to th^ centre of the circle, we have 
(Ara-ia-rt^ - (kx^- p^ = {kx^-p^. 



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therefore l:xi^+2xiSap » Jtaff+2XiS^ « 2rj^ + 2a^Syp; 
hence, eliminatiiig Jk, and making obvious sab- 
Btitations, we get 

S(aa + bfi + cy)p = 0, 
where a+^+c » 0; and therefore the locus is a 
straight line perpendicular to the vector 
f/^w(aa+hfi-¥cy). 

If AB, BC, DO be the directions respectively of 
a,fi,yi and if, arithmetically, AB + BO » CD ; 
then AD is the du-ection of/. 

n. Quaiemhn SoUdion by F. D. Thomsoh, M.A.; and G. A. OaiLYiE. 

It is easy to see that the locus must be a stnught line through the origin, 
since the triangle formed by the moving points always remains parallel and 
similar to its initial position ; but it may be interesting tojgive an investi- 
gation of the locus by the method of Quatemiona, 

Let O be the point of departure ; A, B, 
C the points at the time t; u,v,w vectors 
representing the constant velocities of the 

Stints. Then, writing a, fi, y for OA, 
B, 0C« we have 

« _ ^ ^ 7 . ^^ ^ 

Now, let P be the centre of the circle circumscribing ABC, and let OP=p. 

Then we have to express the fiicts, first, that P is in the plane ABC» and 
second, that it is eqmdistant from ABO. 

The first condition is equivalent to the assertion that PA, AB, BC are in 
one pkne, or that S.PA.AB. BC = 0, or 8.(o-p)(/3-o)(7-/3) -0, 
or S.p(Voi3 + Vi87 + V7o)«So37, or S.p(Vtto + Vtw+ViwO«<S«iw...(l). 
The second condition gives 
(a-p)« - {$-py = (7-pft or o?-2Spa - fiP-2Spfi = 7*-2Sf^. 
or ' 2Sp(iB-a) » iB«-o», and 2Sp(7-/3) = 'f-fiP; 

or 2Sp(o-fi) « t (©»-«'), and 2Sp(wr-w) - <(ftr»-i»=) (2, 8). 

Eliminating t^ 

8.p(yuv + Yvw + yum) _ 2Sp{v'-u) 2Sp(io-p) 
Suvw " t>8-i«« " «»-©» ' 

a system equivalent to two equations, each of the form SpSsO, and therefore 
representing two planes through the origin, their intersection being the locus 
required. 

Similarly, it may be shown that the locusof the intersection of perpen- 
diculars of the triangle ABO has for its equations 

[The three directions of motion are, in the first solution, supposed to be in 
one plane, but in the second solution, anywhere in space.] 



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274S. (Proposed by the Editos.)— Show that the average area of all the 
ellipeea that can be inacribed rpimetrically in a portion of a parabola cut off 
by a doable ordinate perpendicular to the axis^ in parts of the area of the 
parabola, is -^ir if (1) the centres be eqaably distributed over all possible 
positions on the axis, and 4^x if (2) the intersectiona of the chords of contact 
with the axis be eqaably distributed on the axis. 



I. Solution hf Stephbk Watsoit. 

Let AiB'^the doable ordinate, CD the axis, 
PQ the chOTd. of contact cutting CD in I, O 
the centre of the inscribed ellipse, and EG a 
common tangent at Q, meeting CD produced 
in E, and the semi-diameter O^ perpendicular 
to CD- produced, in 0. 

Put CD - a, CB = J, CO - o, OP « /3, 
01 » ox ; then we have 
i^«OE.01-=(2lD + OI)OI-2(a-«)aa:-a«ar«, 

therefore a^a[{2x)^^x}, and EI(-.2DI) : IQ » EC : 0G> 

2D1 " 2a " 2a3jr * 




therefore 



/i»-OQ.IQ«- 



.(A). 



'-M'-fi)'!' 

hence the area of the ellipse hi parts of that of the parabola it 

g = f{(2,)*-.}(l-(|)»} 

(1). In this case the number of posdble positions of O is |a, and the limits 
of # are and \ ; hence the re^uirad average is 

lf^(x)a. . ^f{i2.)^-.} {x-(|)»} {(2.)-»-i}«». . ^ ,. 



(2). In this case the number of positions of I is a ; and the limits of jr, and 
\ I hence the required average is 

i/-(A).(,.<») - ^f {(a,)*-,} {i-(|)»}(2,)-».i« = g,. 



11. Solution by the '^BX^BOSEA, 
Let the aquation to the parabola be 

y^«4ma? (1); 

then, (a, /3) being the semi-axes of an inscribed ellipse, its equation will be 

(frfli^\^.i (2). 



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Eliminating y fh>m (1) and ^2), and applying the condition for equal roots 
to the resultant, we find that the ellipse (2) will have doable symmetric con- 
tact with the given parabola (1) if 

iB«+2ma-5/3, or a - 5^1^ (3). 

2fli 

Otherwise : The equation of a conic which has double contact with the 
parabola (1), along the chord PQ or x—z^O where z « DI, is 

5f«-4ma? = it(a?-«)' (4)5 

and determining Jk so that (4) may touch the double ordinate AB, we find 

k - ^^ ^ 

hence the equation of an ellipse inscribed in the parabola becomes 

anditssemi-axesare a^^z£, ^^^(a + ^) (g). 

2a o 

From (3) and (6) we see that the area of the ellipse, in parts of the 
parabola, is 

gj(i/3«-i8'). or ^(a+»)(a>-»«) (7.8). 

In (1), a varies uniformly from to ^ ; hence, by (7), the average area 
of the inscribed ellipses is 

5^3/* *"(ft/3-i8») d$ » ^y * (&/3-i8») (6-2/3) d» 

i 

In (2), z varies uniformly from to a; hence, by (8), the average area of 
the inscribed ellipses is 



From (7) or (8) we find that the inscribed ellipse is a maximum when 
ziB^a, a»^t fi — if^i its area being )ir. 



2640. (Proposed by the Rev. J. Wolstenholme, M.A.)— 1. If a conic 
of given eccentricity e circumscribe a triangle ABC, the locus of its centre is 

(yz sin' A + ga? sin» B -h a?y sin' C)' ^ (g»- 2)* , 
a?y«(af+y + ») sin' A sin'Bsin'C 1— «' ' 

TOL. XI. 



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nnd (2) if it be inicribed in the triangle, the locus of the centre is . 
(a»8in'A-f...-f2ygginB8inCco8A-f ...y ^ 4(e»~2)\ 
xyz (a? +y + 2) sin* A sin'' B sin* C 1 — c' ' 

the triangle of reference having its angular points at the middle points of the 
sides of the triangle ABC. 

8. Hence show that the locos of points with which as centre two similar 
conies can be drawn, one inscribed and the other circumscribed to the tri- 
angle, is two circles, one the circumscribed circle of the triangle, and the 
other that which has the centre of gravity and the centre of peipeudicnlars 
of the triangle as the ends of a diameter. 



I. SolvHon hff W. S. M*Cat, B.A., and the I^oposeb. 

1. If 2 and A be the tangential equations of a conic and the circular points 
at infinity; then the values of Jfl, for which 2 + Ar^A » represents points, 
are proportional to the squares of the axes of the conic, and are given by the 
discriminating equation 

A' + Ae'F + eit^^O, 

for this is evidently so in the case a^K^ + 6=^'— 1^ + *> (X' + /i>) — 0. 
Hence, easily, if ^ be the angle between the asymptotes 

e being positive for an ellipse and negative for an hyperbola. 
Now for the circumscribiug conic 2 (fsfz+ffxz+hxy) s 0, 
— e =/« sin«A+ ... --Tigh an B sin C..., 
-e'«2(/cosA+^cosB+AcosC). 

But /:g:h^^(^x+Y+z):( ) : ( ), 

smA 
in terms of areal coordinates of centre, 

/:,:*-41|,: £±|y :£±£«. 

sm A sm B sm C 

the coordinates being referred to the triangle whose vertices are the middle 
points of the sides of the original triangle. Houce we obtain at once the 
first equation given in the question. 

2. Again, for the inscribed conic (Za?)* + (my)* + (n«)* « 0, 

e » 2lmn (I sin B sin C -I- m sin A sin C -I- n sin A sin B), 
e'. P + iB« + n«+2»incosA+...., 
and Z : f» : N = sin A (— X + Y + Z) : ( ) : ( ) » or sin A : y sin B : z sin C. 
Hence we obtain. the second equation given in the question. 

3. Eliminating e from the equations in (I) and (2), we see that if with 
the same point (x, y, z) as centre there can be drawn two similar conies, one 
inscribed in the triangle ABC and the other circumscribed about it, the 
point (^, y, z) must lie on one of the two circles 

ar2 giui A + ... + 2y« sin B sin C cos A + ... « ±2 {yz sin' A + ,„\ 
If the lower sign be taken, this is the circumscribed circle of the triangle 
ABC; and if the upper, it is the circle in which th>3 centre of perpendiculars 
and the centre of gravity are ends of a diameter. This latter drcle is there- 
fore one of the system coaxal with the circumscribed and nine-point circles. 



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II. Solutions ly the FbofoSEB. 

1. The equation of a conic circumscribing the triangle ABC, referred to 
AB, AC as axes of Cartesian coordinates, is of the form 

AX(X-c) + /iXY+j^Y(Y-6) = 0; 
and if e be the eccentricity, 

(e3--2)2 ^ 4(X + y~/iiC0sA)2 , 
1-ea 8in2A(4j^X-/i2) ' 

also for its centre X(2X-c) + /iY « y(2r-6)+/iX = 0; 

but if we transfer to areal coordinates measured on . the triangle ahc (the 
middle points of the sides of ABC), 

c 

and therefore for the centre 

Xc^ vK^ __ iiho 

(x+y)e y{x-¥z) 2y«* 
and the locus of the centre, when the eccentricity e is given, is 

(g2_2)' {2(ir + y)sin2B+y(iP + «) sin'C — 2y«sin BsinCcosAJ^ 
!-«" sin'A sm^B sin^C | y« (a? + y}(ar + a) -j/2;jsj 

which easily reduces to the form given in the question. 

2. The equation of an inscribed conic is 



Aft 



With the condition Tt - ^"1 Tv - t^ « -^ ; 

\h cj \k h) Kk 

XT 1 

and for the centre - » — « __ -. ; 

% . k 2(1— X2^ 

and oombimng these, we find 

|.|.sa-..,.|L(f.|-|). 

Transforming as before to the triangle <tbe, we find 

x«. f*-— -. ^(*+Sf)«^(ar + «)=ar(*+y + z). 

{x+y)(x+z) e b 

But, for the eccentricity e, we have the equation 

sin«AJJL«^i=2^n 

{(i?+y)2 8in3B + (a?+z)3 8m2C-2(a?+y) {x + z) mnB sinC cosA 

_. +.4sinB sinC cosAyg}' 

4sin2Asin2Bffln»C.y«.a?(a?+y + 2:) 
mm (*^M^^^A -»- . ... +2ygsinBMnCco8A+ .*..)' 
4ayx (a? +y + «) sin' A sin^ B sin' C 



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Another wlation may be obUuned in the following manner :-* 

1. If the triangle of reference be the triangle having its angles at the 
middle pointe of the ndes of the triangle ABC, the equation of a conic dr- 
cnmscribing ABC is 

Z(«+y)(«+«) + m(y + s)(y + «)+ii(« + a?)(«+y)-0, 
and the equation of the stnught lines through A parallel to its asymptotes is 

whence, if ^ be the angle between the asymptotes, 

tan^^ « «n*Asin»BBin»C {(m-ni-Z)»-4fim} 

{ n sin^ fi — (m + n - 2) sin B sinC cos Af m sin^C }'* 
Now, if e be the eccentridty and (X, T, Z) the centre, 

<f=^'--4cot«^, and ?_ - _? ?_ ; 

l-«» ^' X(Y + Z) Y(Z+X) Z(X + Y) 

whence, if e be given, the locus of the centre is 

l-d" sin'Asin'BsinaC { YZ(Z-f X)(X4-Y)-yZ«} 

(«»-2)« " jz (X + Y) sin«B-2YZ sinB sinC cosA+Y(Z+X) sm»C}' 

^ XYZ(X+Y + Z)rin«Asin8BrinSC 

"~ JYZsinSA+ZXsin'B + XYsin'C}'' 

2. With a nmilar notation, the equation of an inscribed conic bdng 

^j^j . 16 Imn (l + m + n) sin* A sin' B sin' C 

{ r^ sin'A + m2 Bin3B + n>sin>C + 2mji sin B8inCoosA+. .. + ...}'' 

X T Z 

and — s z s -, whence the locus of the centre of an inscribed conic of 
Imn 

g^ven eccentricity e is 

(gg-2)8 (XgsinlA+...-f2YZsinBMnCco8A-h...)a 

l-fl« " 4XYZ(X + Y + Z)smSAffln»Bsin«C ' 



2781. (Proposed by the Rev, J. Wolbtbnholmb, M.A.)— Prove that 
the maximum value of the common chord of an ellipse and its drcle of curva- 
ture, a and b being the semiaxes of the ellipse, is 

I. Solution hif Matthew Collins, BA. 

Let A A' a 2a be the major axis, O the centre, 
and PQ'a ^ =r the tangent PQ; then PQ'R is 
well known to be a chord of the drcle oeculating 
the ellipse at P. Let OMsor, the perpendicular 
MP— y; then we have 



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Again, drawing FQ'B' parallel to PQ, and putting Q'B » z, we have 
QA.QA^ _Q^A.<yA^ Q^A.<yA^ . ^-«' ^ «'--(^~ - 2arj 



a' 



V ar / ^ (ag~a?g)(4a:»~o») 4a^-ga 






a?» 
therefore PB « < + « «^ < ^1 + 1^ - < ^l^y 

Now, when this is a maximum, 

must also he a maximum ; its differential coefficient equated to zero gives 
3(a'-i2)ar*-2o«a!3(2a3-52) + a« « 0, whence — = 2a»-i3+K - 0Q« 

where R « (a<-a2^ + J<)* is plainly > a'-^ftS; and therefore, as 0Q2>aS, 
the proper value of OQ* is 2«'— J2 + r . hence 

»« a«^ ^ 2a«-62+R 

_p _^ 2g«--&«-R _ 2R + 2a»-ft8 . 
8 8 ' 

therefore PR2« 15 <»ar*- 15^' -15^ 2R + 2«»~y 
a* /a*y 8 *(R + 2aS-62/ 

_16a* (2R + 2o»- h^ (2a^ -f &« - R)a 
3 ' {3a2(a3-6«)}* 

"^ 27I^*^^^--®^'('^-^>^(^«'-*'^l 
' 27(a'-ft3)2 {^R'" (2«^-^) [3R^-(2«»-6^)2] 

the required expression for the square of the maximum common chord. 

II. Solution hy the Proposeb. 

The circle of curvature of the ellipse ^ + ^ « 1, at the point whose ec- 

a' or 

centric angle is 0, meets the ellipse again in the point (a cos 36, -6 sin 30), 
and the square on the common chord is therefore 



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«3«48m«26(a»8m«6 + i«co8«e)«2(l-»«)(a« + ft«-A), if c = co82#. 
For a maximam or minimam value, 

either ^-0, or 8c3«'-2(a«+6=)«-c» - 0, 

do 

of which the former gives u^O; and since «' is essentially positive, this is 
the minimum value. The maximum value is then given by that root of the 
latter equation which is numerically less than 1, or by 

a2 + ^,2(a<~aa6a4-&<)* ^ ga.4-y~2R 

For this value we have 

i«2 « ?i J2(a5 + bO^ + c«}- ?lfl!±^^ {2 (a« + b«) « + ca} --2c2fi + 2(a2i-b«) 
d 3c* 

9e> ' ' 8£!» 8 3 

„_^«{(a' + b»)> + 3(a»-b»)'} + ^(a»+b>) , 

- -i?. (a«-a»l)S+M)(a« + 6S-2B)+ ^(aS+JO 

_ ii {2R'-(a'+6'j [(a<-a»6»+M)-3(o'-bO']} 

" WW^ '^^"*~ a>6' + M)5-(o»+ 6») (2a»-6») (26»-a')} . 

The longest common chord has therefore the value given in the question. 



5. (Proposed by H. R. Gsebb, B.j^.) — Determine in trilinear co- 
ordinates the foot of the perpendicular firom the point (a;', y\ s^ on the line 
Ix + my + nz'BiOi also the reflexion of the same point with regard to the 
same Une. 



Solution hy James Dale. 

If the line La?+My + N««0 (1) 

be perpendicular to Ix-hmy -f ns » . . . . • (2), 

and pass through (a/, y, /), we have the conditions 

Lar'+My + Na'-O, W + Mm'-^Nn'^O 
(where i'= Z - n cos B— m cos C, with corresponding values for m', «') ; 

L ^ M ^ N 
therefore I »»' »' I \n' V [^ \ V m' 

\ y' z'\ \z' j/\ I a?' / 



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The intersection of (1) and (2), that is, the foot of the perpendicular from 
the given point on the given line, is 



J_ . 



m n 

M N 



n I 

N L 



f_ 



/ m 
L M 



and substituting for L, M, N, we get 



« 




V 




C 


m n 
n' V V mf 
2! sf ^ y' 




n I 
V m' m' n' 
a// y' ^ 




I m 
y' 7! Uaf 



2a 



a 


b 





I 


m 


n 


m' n' . 


«' V 


V m' 


y ^ 


7! a/ 


X' y' 



The coordinates of the point of reflexion (a?i,yi, «i) may now be found from 
the equations 2{ = a:' + x^, 2ij = y + yi, 2{ = ^ + Zi, 



230L (Proposed by W. K. Clifpobd.)— A circle is drawn so that its 
radical axis with respect to the focus S of a parabola is a tangent to the 
parabola ; if a tangent to the circle cut the parabola in A, B, and if SC, 
bisecting the angle ASB, cut AB in C, the locus of C is a straight line. 



Solution by the Rev. J. Womtenholme, M.A. 

Through the foci S, S' of a conic draw a drcle, and 
from any point T on this circle draw two tangents ; 
the straight lines bisecting the angles between these 
tangents will bisect the angles between TS and TS', 
and will therefore pass through the fixed points where 
the circle meets the minor axis. Reciprocate the sys- 
tem on S; then to the circle corresponds a parabola 
having S for its focus, and to the ellipse a circle having 
for radical axis to it and the point-circle S a tangent 
to the parabola (the reciprocal of S') ; to the point T corresponds a tangent 
to the parabola ; to the tangents the two points A, B ; and to the straight 
lines bisecting the angle, points on AB where the straight lines bisecting 
ASB meet it. Whence the theorem. 




2456. (Proposed by Dr. Shaw.)— Show that if an ellipse pass through 
the centre of a hyperbola, and have its foci on the hyperbola's asymptotes ; 
(a) the hyperbola passes through the centre of the ellipse, {b) the axes of 
each curve are respectively tangent and normal to each other, and (c) the 
two axes which are also noimals are equal to each other. 



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SoUtHon by W. H. Latbbtt, B.A. 
(a ) Let the equation to the ellipse be — + ^ s 1 ; therefore the equa- 
tions to the asymptotes of the hyperbola are 

ay'-y(a?'-c)-c/«0, and ay'-y(»' + c) + cy'= 0, 

where the centre (a/, y) of the hyperbohi is on the ellipse. Hence the equa- 
tion to the hyperlx)la is 

where Xr^ is a constant. The cunre then will not pass through the centre of 
the ellipse unless k=^0, when the equation becomes 

*y»+5^2(^3_ca)_2aya:y + 2c«yy'-0 (1). 

(6.) If, in (1), we put yssO, we have aj'^O; therefore the major axis of 
the ellipse is a tangent to the hyperbola at the ori$;:in, and the m^nor axis is 
a normal; also the axes of (l), being the bisectors of the focal distances, are 
tangent and normal to the ellipse. 

(c.) Again, transform the equation (1) so that (a/,y) may be the origin, and 
the tangent and normal to the ellipse the coordinate axes ; the trai^ortned 

equation will be ? ^ ^ ^ = 1. 

Therefore the axis of the hyperbola, which is a normal to the ellipse, is equal 
to h, and is equal to the minor axis of the ellipse, which, by (b), is the normal 
to the hyperbola. 



2653. (Proposed by J. Gbi7VITH8, M.A.)— 1. If the centre of one of the 
four circles which touch the sides of a triangle self-conjugate with respect to 
a parabola lies on the directrix ; show that its circumference passes through 
the focus. 

2. If the point of intersection of the three perpendiculars of a triangle in- 
scribed in a parabola lies on the directrix ; show that the self-conjugate circle 
of this triangle passes through the focus of the curve. 



Solution Ity the Rev. J. Wolstenholme, M.A. 
1. If we take the triangle as that of reference, the equation of the para- 
bola is 2a^+ fiMf^-k-ns^ » 0, with the condition - + — + ^ -> 0. 

I m n 

The equation of the directrix is x ( — + Sl\ + .... +....«0; 

\n mj 

which will pass through the centre of the inscribed circle, if 



'(t*5- 



+ ....«0, or ^+.... + ....«0. 
al 



Hence J (1+f - ili j _ „ (1±* _ *1^) . , ^112 _ ItiV 
or la (A— c) « mb {c^a) ^ nc (a— b). 



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But the foeos is giren by tbe eqnataons 

— -r — .-.,.., or I 
•i6«+nc» b ^ c 



c—a a—h 



and this point manifestly lies on the inscribed circle («cot j A/ + ... + ... »0. 
This point is of course the point of contact of the inscribed, or escribed, circle 
with the nine-point circle. It may be noticed that the tangent to this circle 
at the focus of the parabola is the polar with respect to the parabola of the 

centre of the circle, its equation being -^— + -i^ + -f— » 0. 

h—c c^a a^b 
There is no difference in the work when one of the other circles is taken, 
beyond changing the signs of one of the three, a, 6, o. 

2. This is only the reciprocal of the above proposition with respect to the 
focus of the paraboU. 



2718. (Proposed by Professor Caylby.) — Find in piano the locus of a 
point P, such that from it two given points A, C, and two given points B, D, 
subtend equal angles. 

2767. (Proposed by Professor Cayley.)— If 



*o'+yo" = l, 



la?, y, 1 I « L ; 



show that each of the equations 

ag(x-Xoy + &g(jr->.ye)« ^ a«(a?>ar,y + &a(y-yt)« ^^^^ 

(a?aro+5(yo-l)* (a^ari+yyi-l? 

g»(a?~aroy-|.6«(y-yoy ^ (^(x-Xi)^ + hHy-y{)^ /g)^ 

(ajyo-»oy)^-(*-'o)*-(y-yo)' (a?yi-a?iv)«- («-a?i)*-(y-yO»"* 
represents the right line L « and a cubic curve. 

1819. (Proposed by C. Taylor, M. A.) —From two fixed points on a 
g^ven conic pairs of tangents are drawn to a variable confocal conic, and 
with the fixed points as foci a conic is described passing through any one of 
the four points of intersection. Show that its tangent or normal at that 
point passes through a fixed point. 

Solution of the above Problems by Pbofbssob Caylby. 

I. It is easy to see that drawing through the points A, C a circle, and 
through fi, D a circle, such that the radii of the two circles are proportional 
to the lengths AC, BD, then that the required locus is that of the inter- 
sei^tions of the two variable circles. 

Take AC « 2Z, MO perpendicular to it at its middle point M, and ^p; 
a, b the coordinates of M, and A the inclination of p to the axis of a; ; then 
coordinates of O are a +>» cos A, b+p sin A, 
coordinates of A, C BTea±l sin A, d + / cos A, 

VOL. XI. D 



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and hence the equation of a circle, centre O and passing through A, C, is 

(x-a-p C08X)'+ (jf—b—p sin X)« « P+i?*; 
or, what is the same thing, 

(»-a)«+(y-&)'-l« « 2p[(a?-a) cos X+(y-J) sin x]. 

If 2m, 9, 0, <{, ^ refer in like manner to the points B, D, then the equation of 
a drde centre say Q^ and passing through B, D, is 

(»-c)»+ (y-d)»-iii^'- 22 [(a:-c) cos ^ + (y-d) sin /«]. 

And the concUtion as to the radii is I? +p» : nfi-^f^t- : m^ that is, j^ : ^2= 
r^ : ffi', or p : j[ » + / : »i. And we thus have for the equation of the re- 
quired locus 

(x-a) cos X -»- Cy-6) sin X " "• m (4f-c) co8/A + (y-(0 sin /i' 

▼iz., the locus is composed of two cnhics, which are at once seen to he 
circular cubics. One of these will however belong (at least for some posi- 
tions of the four points) to the esse of the subtended angles being equal, the 
other to that of the subtended angles being supplementary ; and we may say 
that the required locus is a circular cubic. 

2. If two of the points coincide, 
suppose C, D at T; then, taking T 
as the origin, we may write 

a a { sin X, ft » —2 cos X, 

CB^msin^, d*- mcos/i. 




and the equation becomes 

or +y^a' 21 {x sin X— y cos X) 
jT cos X -f- y nn X 
_ , i g^+.y' •»■ 2m {x sin fu—y cos /i) 
" m xoos/A+ysin^ 

▼iz., this is 

(as^+y*) [m (a? cos /i +y sin /i) + I (a? cos X +y sin X)] 

^2lm{{x sin X— y cos X) {x cosfi-^y sin /i) 

+ (x sin /A— y cos fi) {x cos X +y sin x)^ = 0. 
Taking the lower signs, the term in { J is (ac^+y'^) sin (X— /i), and the equa- 
tion is 
(«» + y3) ^„| (ay cos /A 4-y sin /i) + Z (a? cos X +y sin x)— 22m sin (X— /*)} =0, 

viz., this is rr' + y' = 0, and a line which is readily seen to be the line AB ; 
and in iact from any point whatever of this line the points A, T and the 
points B, T subtend supplementary angles. 

Taking the upper signs, the equation is 
(a* +y*) [w (a? cos /i +y sin /:*)— Z (ar cos X +y sin X)] 

-2^{(a^^-y3) sin (x + fi)-ary co8(X + m)} = 0, 

which is the locus for equal angles, a circular cubic as in the case of the four 
distinct points. 

8. The Question is connected with Question 1819, which is given above. 

' In fact, taking A, B for the fixed points on the given conic, and P for the 

intersection of any two of the tangents, if in the conic (foci A, B) which passes 



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throngh P, the tangent or normal at P passes through a fixed point T, then 
it is clear that at P the points A , T and B, T subtend equal angles, and conse- 
quently the locus of P should 'he a circular cubic as above. The theorem 
will therefore be proved if it be shown that the locus of P considered as the 
intersection of tangents from A, B to the variable confocal conic is in fact the 
foregoing circular cubic. I remark that the fixed point T is in fact the in- 
tersection of the tangents AT, BT to the given conic at the points A, B 
respectively. 

4. Consider the points A, 6, (which we may in the first instance take to be 
arbitrary points, but we shall afterwards suppose them to be situate on the 

conic — + ^ » 1,) and from each of them draw a pair of tangents to the 
conibcal conic \- -^L. » 1. Take {xq, yd) for the coordinates of A, and 

<r + h 0* + h 

(^ yd ^^^ those of B ; then the equation of the pair of tangents from A is 

or, what is the same thing, 

(a«tA) (6«+A)" a« + A "" b« + A ' 
that is, (a!yo-*oy)*-('f>*+*) («-afo)'-(»*+*)Cy-yo)*=0, 
or as this may also be written 

and nmilarly for the tangents from B we have 

(ayi-ari|/)a-b3(a? ^x{f-a» (y-yi)« = h [(jr-afi^a + Cy-yO*] 5 

in which equations the points {xq, yo)» {^i* yi) ai^ in &ct any two points 
whatever. 

6. Eliminating h, we have as the locus of the intersection of the tang^enta 

(ayo-gbyy-&' (x-xo)^-a? (y-yp)' ^ (ayi-a?iy)'-^(g-a?i)^~«'Cy-yi)' 
(a?-aro)« + (y-yo)' (a?-a?i)2 + (y-yi)« ' 

which is apparently a quartic curve ; but it is obvious d priori that the 
locus includes as part of itself the line AB which joins the two given points. 
In fiict, there is in the series of confocal conies one conic which touches the 
line in question, and since for this conic one of the tangents from A and also 
one of the tangents from B is the line AB, we see that every point of the 
line AB belongs to the required locus. The locus is thus made up of the 
line in question and of the cubic curve. 

6. To effect the reduction it will be convement to write ax, "by in the 
place ofx,y, (outo, b^g, axi, byi in place ofx^y^ x^ yi,) and thus consider 
the equation under the form 

a«(ar-aro)»-f-l>«(y-yo)« ^ aH'-^xf-^^ {V-Vi? 
(ayo-'qy)*— («-a?o)'-(y-yo)* (a?yi-a:iy)^-(a?-a?i)«-(y-yi)» 

It is to be shown that this equation represents the line L«0, and a cubic 
curve. 



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36 

Writing for a moment «o-» + 6». Vo-y + ito. andiTi -» + |, y, -y + iH, 
the equation becomes 

and hence, multiplving out, the equation is at once seen to contain the 
factor ^i|i— liiio (which is in fiict the determinant just mentioned), and 
when divested of this ikctor the equation is 

-b« [(y"-l) (&i?j + 4ii?o)-2aryiyoi?i]. 
Writing herein for ^q, %, |,, iy, their talues, and consequently 
&€i - *■-» (aro+a^i) +«o»i . 

loiJi + fii?© - 2apy-ar(yo+yi)-y (aro+a?i) + arQ5fi + fl?iyoi 
and arranging the terms, the equation is found to be 
(a«af«+6»y») ["a?(yi + yo)-!/(*i + aro)] + (a»*« + bV)(a^ayi + ^a'o) 
- 2ary [a« (1 + a?oari)- b» (1 + yoyO] 
+ (a«-b«) [a?(yi + yo) + y (a?i + a?o)-(aroyi + a?iyo^] - 0, 
which is the required cubic curve. 

7. Restoring the original coordinates, or writing -, ^, ??, &c. in place of 

aba 
*» y> ^'o, &c., we have 

(*' + y*) C-a^ (yi + yo) + y (a^i + aro)] 

+ (a?»-y8)(aroyi + *iyo)-2a?y {a^-l^ + Wffti-yia\) 
+ (a2_ba)[flj(yi + yo) + y (ari + aro)-(a?flyi + ariyo)] = 0, 
which is a circular cubic the locus of the intersections of the tangents from 
the arbitrary pomts (a?©, yo), (a?i, yO to the series of confocal conies 
. X* y* 

o^Ta "*" 6«Ta "^ ^ ' *^® origin of the coordinates is at the centre of the 
conies. 

8. Supposing that the points (aro, yo)» (*i> yi) are on the conic 
^ + I, - 1, and that we have consequently ^% ^ « 1, 5l! +^ = 1, 
the equations of the tangents at these points respectively are 

and hence, writing for shortness a=yo-yi, /3«ari-afo, 7-aroyi-a?iyo. we find 

* "TT'^'*"" — as the coordmates of the point of intersection T, of the 

7 7 

two tangents; and in order to transform to this point as origin, we must in 

place of », y write ar- ^, y- ^2i respectively. Or what is more con- 
venient, we may m the equation at the end of (6), in which it is to be now 

as8umedthata;e2 + yo«=.l, *|« + y,««l, write «-^ y ^ 5 for *, y. and 

7 



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then restore the original coordinates by writing -, ^, ^, &e^ for «, y, ttf^, Ac, 

aha 

and f , ?-, JL foi a, ^ y, these quantities throughout signifying a »yo— yi, 
h a ah 

fi s a^— jTo, 7 ss a?oyi— *iyo« I howerer obtained the equation referred to 
the point T as origin by a dUferent process, as follows:— 



9. Starting from the equation at the commencement of (5), I found that 
tie points (xq, y©), (a^i» yi) lacing od 
could be transformed into the form 



the points (d?o>yo)> (^i>yi) being on the conic ^ ^V-^ ^J, the equation 

a* o* 



an equation wluch (not, as the original one, for all valoes of (xq, y^, (4^, ^i), 

but) for values of (jto, ^o), («i, yO* »«ch that f^ + ^ = l, ^ + ?l' „ 1, 

a^ or a* 0^ 

breaks up into the line AB and a cubic curve. 

10. To simplify the transformation, write as before ax, by, tueo, &c., for «, 
y, dPo» &C. We have thus to consider the equation 

(a?aro+yyo-l)' («?«?! +yyi-l)' ' 

where xf-\-y^ » 1, x^+pi^ — I9 and which equation, I say, breaks up into 
the line L»0, and into a cubic. 

Write for shortness a = yo— ^b fi ■■ *i— fl?o, y = fl?Qyi-"«iyo» so that the 
equation of the last mentioned line is av + /3y+7»0. Then it may be 
verified that, in virtue of the relations between (xq, yo), {xi, pi), we have 
identically 

(a?-a?o)(*ari+yyi-l) + (a?-ari)(araro+yyo-l) = (aX'¥fiy + y)^^SdLEl(yx-i-a), 

(a?— aro)(*a?i+yyi— 1)— (a?— a?i)(asro+yyo-l) *= /8«*— oaiy— 7y-iBj 
and, fflmilarly, 

(y-yo)(a?a?i+yyi-i)+(y-yi)(*'o+yyo-i) -(«*+/3y+7)?^* (7y+/3). 
(y-yoXa^ari+yyi-^-Cy-yJCipafo+yyo-l) = i3ay-ay»+7x+o. 

11. The equation in question may be written a'F + ft'Q » 0, where 

P « (a?-aro)'(a?a?i+yyi-l)'-(a?-*i)'(«»o+yyo— 1)'» 
Q - (y--yo)'(a?ari+yyi-l)«-(y-yi)2(ara?o+yyo-l)^ 

values which are given by means of the formula just obtained ; there is a 
common factor ax + fiif-f-y which is to be thrown out; and we have also^ as 
is at once verified, ^(2JL^ » xp + xi^ ^ ^^^^ iheae equal Victors may be thrown 

out. We thus obtain the cubic equation 

a»(7« + a)(/to*-cay->y-^) + 6'(7y + /3)Oay-c8^ + 7J? + o)-0. 



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This is nmplified by writing d? — ~ for d?, ^— - for y. It thne beoomei 

y 7 

aVr [(7«-a) (/8*-ay)-7»i^] + b'y [(7^-/3) (iB*-<v) + 7*'] = Oi 
or, what is the same thing, 

thatis, 7(«'*' + *y)0to-ay) + a»[-a^«« + (a»-y)ary] 

12. Eestoring f . S, fj for a?, tt^ x^ and 1?, ?!?, «> for y, yo, yi I 

a a a a a a 

writing, consequently, ^, r, X in pkce of a,fi,y, \£ a,0,y are still used 
a (to 

to denote a » yo-Vi* iB ■■ a?!— ^^o. 7 • *oyi— *iyoi t^® equation becomes 

+ b»[-(b^/5»-7^ay + a^a^1/^] = 0, 

where now, as originally, ?^ + ?^ = l, ^ + 5^ — 1; viz., this is the 

equation, referred to the point T as origin, of the locus of the point P con- 
sidered as the intersection of tangents from A, B to the variable confocal 
conic; and it is consequently the equation whidi would be obtained as indi- 
cated in (8). The locus is thus a circular cubic ; the equation is identical in 
form with that obtained (2) for the locus of the point at which A, T and 6, T 
subtend equal angles, and the complete identification of the two equations 
may be ^ected without difficulty. 

13. I may remark that M. Chasles has given (Comptes Rendus, torn. 68^ 
February, 1864) the theorem that the locus of the intersections of the tan- 
gents drawn from a fixed conic to the conies of a system (^, y) is a curve of 
the order 3y. The confocal conies, qud conies touching four fixed lines, are a 
system (0, 1) ; hence, tajcing for the fixed conic the two points A, B, we 
have, as a vcoy particular case, the foregoing theorem, that the locus of the 
intersections of the tangents drawn from two fixed points to a variable con- 
focal conic is a cubic curve. 



2778. (Proposed by Professor STLYEdTEB.) — Suppose /a any quantity 
fracUonal or integral, and write 

also let n be a positive integer such that /i and ^+fi have not different 
signs ; then it has been proved elsewhere, and may be assumed to be true, 
that the number of changes of sign in the series 

is an inferior limit to the number of imaginary roots in the equation 
cbir" + Cia»*''^ + Ca»*"'+.... —0. 



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1. Beqirired to prore that the ahove theorem renuuns tme when we write 

r »(' + !) jr(ir+l) (»'+2) 

prorided r is a poatiTe integer, and ^c— r + 1, /i + r do not ^QiFer in sign. 

2. Afloertun whether w hdng an integer ia esaentaal to the troth of the 
extended theorem. 



SolmHoM by (he Vbjofobkr, 
Let /r = e^ + Ci«^~*+...., Fap- ar'+^^+^c, 

1 .2 1 • 2 • 3 

the alternately appearing and disappearing fiictor $ being any quantity Caf^ 
ferent from oiuty, c an infinitesimal of like rign with c^ r any positiTe in- 
teger, and ft. any qoantily (fractional or integer) such that /t, and ^c-f »-|-r do 
not differ in sign. 

^ '•^ 1.2.. ..r "^^ ^" 1.2.. ..(r+1)^* 

Then, by the Postulate of the question, the number of imaginary roots in 
"Fx is not less than the sum of the changes of sign in the two series 
€«5 (tf«-l)€*; (l-fl»)c?; ....; (....); a?; 
a«; af-ooos; ....; a*. 

In each of these series, the term a*, properly speaking, has attached to 
itself another term^ but which may be rgected as of only infinitesimal magni- 
tude. In the penultimate term of the upper series, the parentheris left Tacsnt 
(gimilarly neglecting an infinitesimal term of higher order than the one re- 
tained) will be — c^'^oo <^ — ^'~^ao, according as j^ is even or odd. Thus 
the number of changes of sign in this series is r if r is even, and r— 1 if jr is 
odd. 

Again» the roots of Vx consist of a group of infinitely gpreat and a group of 
finite magnitudes. The former are the roots of the equation dPfar+c^ <= 0» 
or, which is ultimately the same thing, oftx' + CQ'^O; the latter ofjx = 0. 
The number of imaginary roots of the infinite group is therefore r or r— 1 
acoor^ng as r is even or odd ; t. e., is the ssme as the number of changes of 
sign in the upper series. Hence the number of imaginary roots belonging to 
fx^O cannot be less than the number of sign-changes in the lower series; 
but this last number will not be affected if we multiply each quantity, 

a^aj,a^ by the common fiictor A^O*+l)»»»>(M+y-l) ^ or, which is 

the same thing, if we write 

r + l (v+l)(i' + 2) 

Let now r+l*/, /A + r*/, so that 

|i«|i'— r -«/—/+ 1, and /i + n + r «■ f*'+ «. 
Then, suppressing the accents, we see that, » being any positive integer, 
and /I not iutermeduite between r— 1 and — «, if 

c^j^oo; ei^^an c j = ^i^i±i) a , ; .... 



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the number of imaginary roots infx is not less than that of ngn-changes in 
''o*; o^—aQ<H* «a?— «iOs5 •••• 

When v»l and fi^^n^ we &11 back again npon Newton's mle. One 
feels strongly inclined (on more than one ground) to suspect that v need not 
be integer, but I do not see my way to a proof. 

As a corollary, it may be noticed that the equation 

^»+^a;»-l+M> + l)^-2+ 

r r (i' + 1 ) 

or more generally, the real function (where 0^ does not exceed unity) 

V y^y + i) y{y + l){y + 2){y + d) ) 

\y ^y(y + l)(ir+2) / 

when r is a positive integer, and /a— r + 1, /i-fn have the same sign (i. e. 
when /A is not intermediate between r— 1 and — n), cannot have more than 
one real root. 

N.B. — It is worthy of notice, that if we make fx become the middle part 
of a function of x having for its roots a group of infinitely greats, a group 
of infinitely smalls, and the roots of fx itself, we may, by a method of de- 
monstration similar to that here employed, pass direct from Newton's rule 
(as given by Newton) to the final extension of it contained in the question 
above, but with the difference that, so far as the force of such demonstration 
takes effect, fi as well as p will have to be integral, a restriction we know to 
be superfluous. This is one of the reasons for doubting the necessity of i' 
being an integer. 

For the proof of the Postulate, see the Proceedings of the London Mathe* 
matical Society, No. II. The above demonstration would have been just as 
good if only the first powers of the infinitesimal €, instead of the ascending 
scale, had been used in <^ throughout ; but I have not thought it worth 
while to disturb the text by making this simpUficatioD, which the reader can . 
do for himself. 



To jmd the number of permutations of n things taken r together. 
By C. R. RIPPIN, M.A. 

Let P (n, r) denote the number. Now, if » things be arranged in all possible 
ways r at a time, each will stand first as often as the remaining »— 1 can be 
arranged r— 1 at a time,that is, P (» - 1, r— 1) times ; but the whole number of 
permutations is secured by making each thing stand first as ofben as possible; 
and therefore, since there are » things, we must have 

P(w,r)-«.P(n-l,r-l). 
Similarly P(n-l,r-l) = (n-l) . P(n-2,r-2), 

&C. =: &C. 

P(«-r + 2, 2)«(»-r + 2).P(n-r+l,l)«(n-r+2)(»-r+l.) 
Multiply and simplify, then we obtain 

P(n, r) = w(w-l)(w-2)....(«-r+2)(»-r+l). 



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41 

2798. (Propofied by J. J. Walexb, M.A.)— To find the ocmdition that a 
given point (a?, y, z) may be exterior to the given conic 

S = (A, B, C, P, G, H jar.y, 2)' « 0. 

Solution hy F. D. TnoHBOir, M.A. 

If a point be without a conic^ its polar cats the conic in two real points. 
Now let \x ■\' f»y •{' vz — be the equation to polar of x^ y, z. 

Then, for the points of intersection of the polar with the conic, we have 
the equation 

(A, B, C, P, G, H^-(fiy+i«), Xy, \zf - 0. 

Expanding this as a quadratic in £, and writing down the condition for 

z 
real roots, we have 

(F2-BC, G«-CA, HS-AB. AP-GH, BG-PH, CH-FGjx,/u,ir)' > 0; 

(2S c^S c^S 
or, Binoe x : /i : y - ^ : ^ : ^, 

ax dy dz 

(A'.B'.C.F'.G'.H'X|. |. f)'>0 (1). 

where A^ B% &c. are the inverte coefficients. 
The condition (1) reduces to 

(AF« + BG«+CH«-ABC-2PGH)S>0, or AS > (2). 

In the particular case where |/»««0, S «» ksfi and is positive; therefore 
the condition becomes A > 0. 



1012. (Proposed by Professor MAiTErHBiH.)— If a and I be the points of 
contact, with a curve of the third class, of a double tangent ; and if this 
tangent be intersected in 4»» n, p by the three tangents to the curve which 

can be drawn from any point in the plane, then f^'^'^^'f^ » ^, where 

hm,on,})p ph 

pQ^ Pb "® ^® "^ ^ curvature at the points a and b. 

Solution by the Rev. J. Wolbtenhouce, M.A. 

Take {x, y, z) for tangential coordinates, and («, $, y) for areal coordinates. 
The tangential equation of a curve of the third class, to which a==:0 is a double 
tangent at the points where it meets /3»0, 7»0, is of the form 

an/« + A3^ + By»s + Cyx«+D«»-0 ..............(1). 

For the tangents through an arbitrary point (jp, g, r) we sh%ll have 

px + qy-t-rz^^O (2) , 

VOL. XI. E 



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wbicb» oomlnuod with (1), leadi to the eobie 

i>(Ay» + By»t + Cy«*+D«>)-y«{5^+r2) -O; 

to tbat }(lSl^ « 5 and b independent of (p , q, r). But the tangent 

«l2,Z, A 

x^a+ififi + Xiy^O meets the straight line a ■ in the pcnnt where 
,;,+,.,-0. or f?-«;, whence ^^ l.«».t«.t. 

To inye^-tigate this constant, take a 
point P near a, from whidi draw 
three tangent?, and let Z Psm » 8^, 
I^ » 8^\ Then the constant re- 

qnired is the Umit of ^-'^^'^ when 

M moTCB np to h, and n and |? to a, 

- hunt of - — - — J21. 

« ' limit of ^*-°P; 

-A lunitof ^*-^P; 
Pi, ^M' 

and making P approach its limit in such a cUrection that an^^ap^ we shaU 

therefore have ultimately — ^ -> 1, and ap » ip^^^' a -^. 
an p^ 

[A solution by Professor Cbexoha has been given on p. 88 of Vol. YII. 
of the Meprint,'] 




and ^^^^± 
S^ 8^' 8^' 



2740. (PlDposed by W. S. BirBNSiDS, M.A.)~Prove that the envdope 
of a line given by an equation of the form 

Loos2^ + Mnn2^ + Pcos^ + Qnn^ + R»0 

may be obtained as the discriminant of a cubic equation, and form this 
equation. 

Solution hy the PR0F08EB. 

The above equation is plainly reducible to the form 
{abcfgh) (cos^,sin^,l)> « 0, 
which ftirther is equivalent to (abefyh) ft, i|, 0' " or U«0, with 
C + ij*— f — or V=sO; whence, if we find the condition that the conies 
U and y touch (conndering a b efg A as constants), we shall have the equa- 
tion of the envelope required. And this is most easily done by forming the 
discriminant of U + XV » 0, w. (ABCD) (1, X)*, and then equating to zero 
the discriminant of this cubic. 



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263d. (Propoeed by Professor Stltbstsb.)— 

1. If A, B be two confbcal conies, and A' be a third oonic having double 
contact with A, show that there exists a fourth oonic B' having doable con- 
tact with B and confocal with A'. 

2. If A, B be two confocal qoadrics, and A' be a third quadric having 
continuous contact with A, show that there exists a fourth quadric B' having 
continuous contact with B and confocal with A'. 



SoUtHon h^ the Rev. J. Wolstbkholub, M. A. 

1. Generalized by projection, this amounts to : Given two conies A, B 
inscribed in the same quadrilateral (of which 00' is a diagonal) ; there can 
be described two conies A', B' having double contact with A, B respectively, 
and such that the tangents to the two from O, O' are the same ; or (to ex- 
press it more clearly) two of whose common tangents intersect in O and the 
other two in O'. Beciprocating this, we have two conies a, b passing 
through four points lying on the straight lines o, o\ and the theorem to be 
proved is that two conies a\ h* can be formed having double contact with 
a, h respectively and meeting the straight lines o, cf in the same points. 
The simplest case to which this general theorem can be reduced by projection 
appears to be: Given two conies A, B similar and umilarly situated, two 
circles can be drawn having double contiict with A, B respectively, and meet- 
ing their common chord in the same two points. This is not at all hard to 
prove; but as I have not seen how to get a neat solution, I refrHin. 



(Proposed by H. B. Gbbeb, B.A.) — Given a cubic curve K, and 
a ix>iut on it p ; through P is drawn any transversal meeting K attain in m 
and », and on it is taken a point x such that the anharmonic ratio (jpxmn) 
shnll be equal to a given quantity. Prove that the locus of x is a quartic 
curve with a point of osculation at j7, touching E at ;? (counting for four 
points of intersection) and in four other points. 

SoUdion hy James Dalb. 
Beferring E to rectangular axes passing through p, its equation is of the 
form Aaj»+Ba?^+CayS+iy + E«« + Pay-t.Gy2 + Ha? + Ey = 0, 
which, being transformed to polar coordinates, becomes 

r»(A cos? a + B cos* 6 sin 6 + C cos sin* + D sin» 0) 
+ r(Eco8»a + Pcos^sina + Gsiu« «) + Hcosa + Esina «= 0. 

Let now a? be a point on any radius vector cutting the curve in m and ii, 
so that the anharmonic ratio » 2 : m; then, putting p«=r, />»i=ri, pa^rj, 

we get r {(Z-*») ('•i + r,) + (i + fii)(ri-r8)J « 2 (i-m)rir2. 

Substituting the values (ri + ra), (n— •'s)* •'a^'a* obtained from the equation to 
the curve, and returning to x and y coordinates, we get 



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44 

o the equation to the locns of x, which is eyidestly of the fourth degree. 

Let Edr* + Fiy + Gy* + 2 (Hd? + Ey) « 0, which is the eqnalaon of the polar 
conic of p with respect to E, he represented hy P^; and let Ha?+ Ky » 0, 
which is the common taneent at pi to P and K, he represented by T|, and 
the cuhic by Eg; then the locns of » may he written 

«mP,«-.2(Z+m)9Ti.E3; 

and it may he shown that this cnnre touches Eg in the points common to 
T] and P2, that is^ in two coincident points at the origin, T| being the common 
tangent to the three cnrveSf and in fonr other points. 



2494. (Proposed by Professor Etbbstt.)— On each side of a hexagon as 
base, a triangle is described by prodncing the two adjacent sides to meet, and 
a second hexagon is formed by joining the vertices of these triangles in order. 
Show that if either of these two hexagons can be inscribed or circnmscribed 
to a conic, the other can be circumscribed or inscribed to a conic. 

2513. (Proposed by Professor Eyibett.)— Let A, B, C, D, E, F he six 
points. Let BF, CD meet in a ; AB, CE in i ; AB, CD in c ; AF, CE in 
d; BF, DE in ; and AF, DE in/. Show that, if either of the hexagons 
•ABCDEF, abcdrf can be inscribed or circnmscribed to a conic, the other can 
be circumscribed or inscribed to a conic. 



(Proposed by Professor Eyebett.)— If three conies cut one another 
so that every two of them have a common chord which is also a chord of a 
fourth conic^ the other three chords of intersection of the three conies meet 
in a point. 

Solution by the Pbofoseb. 

Theobeu I. [Quest. 2539] Let S be the fourth conic (». e,, let its equation 
be SssO), and a, jS, 7 the three chords common to it and to the others two 
by two; then S— /JB7, S— wyo, S— najS will be the three conies. By sub- 
traction between the two last, we have mya—na$ or a [ ^ — 7V denoting 

a pair of lines which must be chords of intersection of two of the three conies; 

and it is obvious that _-.!, ~ — -» -— — meet in a point. 
m n H II m 

Theobeu II. If the three conies break up into pairs of lines, the foregoing 
theorem has in general, for any given values of a, iS, 7, eight distinct appli- 
cations, that is to say^ there will be eight points in each of which three lines 
meet. 

There are two values of I which make S— l^ break up into a pair of lines. 



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For instanoe, if AC and BD are /3and>, 8—^ may 

denote ather AB and DO or AD and BC. We have thus 

ax piurs of lines which we may denote by 

8 — Ifiy, 8 — mya, S - najS, 
8 - r/iy, 8 - m'-yo, S - n'a^ 

and eight points of meeting, yiz., 




TLQc 1. 



^^l^y. «»A«t; "^^^H, &c, 

which we may denote by Imn, Vm'n', Vmn, Im'n^ Imnf, lmfu\ Vmn\ Vw!n ; 
and the piurs of lines may in like manner be denoted by I, l\ m, m\ n, »'• 
Each of the lines passes throngh two of the points. 

Let A, B, C, D, £, F be any nx points on a conic, and let a, ^, 7 denote 
AD, BE, OF respectively; then 

; is BC and £F, m is CD and FA, 11 is DE and AB; 

r is BF and CE, nl is FD and AC, n' is BD and £A. 

J. — 2 will denote the line joining the intersection of AB and CD to that of 
m % 

FA and DE, which we shall indicate by the notation ''AB, CD to FA, DE." 
The eight sets of lines are 

AB, CD to PA, DE ; BC, DE to AB, EF 5 CD, EF to BC, FA^ 

AC, BD to DP, EA J BD, CE to EA, PB ; CE, DP to FB, AC 

AB, CD to PA, DE 5 AB, CE to DE, BP; AP, CE to CD, BP 

BC, DE to AB, EF; BC, DP to EF, CA ; BA, DP to DE, CA 

CD, EF to BC, FA; CD, EA to FA, DB ; CB, EA to EF, DB 

AC, BD to DP, EA ; EF, BD to BC, EA; EF, AC to BC, DP 

BD, CE to EA, FB ; PA, CE to CD, PB; PA, BD to CD, EA 

CE, DP to FB, AC; AB, DP to DE, AC ; AB, CE to DE, PB 



5 



Vm'n 
Vmn' 
Im'n 
Imn* 
Im'n 
Vmnf 
Vm'n 



The figares of the first two cases are subjoined. In both of them, as here 
lettered, the lines meeting in a 
point are OL, HM, and EX. 
It will be observed that the 
hexagon OHKLMN is formed 
from ABCDEF, in Fig. 1, in 
the same way as ABCDEP is 
formed from GHELMN, in 
Fig. 2. A similar relation con- 
nects ^mn with toV, ^m'nwith 
Vnai^ and ftim' with Vm'n. The 
following figure, in which the 
conic may 1m supposed to pass 
either through A, B, C, D, £, 
F, or through a, 6, c, (f, e,/, 
applies to the two cases Vmn and lm'n\ the capital letters being applicable to 
the former and the small letters to the latter. In 
the former case, <u2, 60, cf will meet in a point ; 
in the latter, AD, BE, CF. We have thus proved 
that, if either of the two hexagons (mentioned 
in Questions 2494, 2513) can be inscribed in a 
conic, the diagonals of the other meet in a point ; 
whence it follows, by the converse of Brianchou's 
theorem, that this other hexagon can be drcum- 
scribed about a conic. 




na. 2. 




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46 

JITofe.— Each of the eight sets of lines is connected with a Pascars line hy 
the relation that, if two triangles be snch that the lines joining corresponding 
vertices meet in a point, the three intersections of corresponding sides will 
be in the same straight line. In Fig. 2 the two triangles are GKM, LNH. 

It will be found that the line E^l is the Pascal ABEDCF, and that £^1 
m M m n 

is the Pascal DBEACF. 



210L (Proposed by M. DAEBorx.)— Tronver los conditions n^cessaire* 
et snffisantes pour que les qnatre racines d'nne Equation du quatri^me degr^ 
forment nn qoadrilatdre inscriptible. Tronver la surface et le rayon de ce 
qnadrilatdre. 

Solution hy J. J. Wauosb, M.A. 

If (a, 6, c, <{) be the fonr real and podtive roots of a biquadratic s^^px^ 
+ 90^— rd?+« s 0, it, is evident that the condition of its being possible to 
oonstmct a quadrilateral figure the four sides of which shall be represented 
in length by these roots is, simply, that the greatest of these values should 
be less than the sum of the other three; t. e.^ that the ezpresnou 

u — (a + 6 + o— d)(a + J— o + ci)(a— & + c + <i)(— a + J + c + d) 
should be poative. In terms of the coefficients of the biquadratic 
tt = 22a»i«-2a* + 8aicd = 43a»i^-(5a9)« + 8aJcrf 
= 4(2»-2pr+2f)-(|^-25)a+8* -i?«(4ff-;^ + 8(2*-pr). 
This condition being ftdfiUedy let a, & be two adjacent sides of the figure, B 
the angle between them, and 8 the length of the diagonal joiniug their ends; 
and let ^ be the angle between the remaining Bides c, d. Then, if the figure 
is to be inscribable in a circle of radius B, we have 

coge + cose'=0, whence JP = (!!i±^4±M±^!)£*; and 

ab-^cd 

cos a - * 11*--^LJl, whence an^ B = Ji L — L .^ Li 

2(aA + c<f) 4(aJ + cd)« 

^ 4(a& + cdP 4(ai + cd)2 

The area of the quadrilateral wm\{ab-k-cd) sin B 

-j{p2(4^-p«) + 8(2*-pr)} (2). 

»3 ((a2 + 62)cd+(c2 + dS)flf6] (ah-k-cd) 

This numerator is a symmetric function of a, &, e, J> being equal to 
:i(^hcd + :ia^lrf^ « (p' - 2g) # + r» - 2gs « (jp^—Aq) « + r^, which is, plainly, 
essentially positive. Hence we obtain 

Ra_ (p'-4g)> + r> (3). 



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47 

In the case of a convex quadrilateral, then, the only condition to be ful- 
filled by the real and positive roots of the biquadratic is «=0; and snch a 
figure is always inscribable in a real circle by giving its angles the values 

determined by sin ■■ —r^— — r. 
2(ffJ + cd) 

If the quadrilateral is to be skew or twisted, 

jj^ (g8+y)c<g-(ca-fd»)g5 
cd—ab * 

and that this should be positive, {(a^+ 6^ cci— (c'+ d*) oi] {fsd^ab) must 
be positive; or Sa^^^c*— Ja'Jcc?, i.e., r*-25*— (/i'— 2^)«, or r*— p*#, must 
be positive. If this condition and «»0 are satisfied, it will be found that 



2616. (Proposed by C. W. MebeitibIiD, F.R.S.) — Let two intersecting 
tetrahedra have all their edges bisected by the same system of Cartesian 
axes, each axis through two opposite edges of each tetrahedron; them the 
solid about the origin has the origin for its centre of figure. 



Bohaion by the Pboposxb. 

The origin is of course the centre of figure of each tetrahedron separately. 
Moreover each tetrahedron is completely defined by the angles betwem tiie 
axes, and the intercepts of the edges, except that it may be inverted. 

Calling the intercepts a, b, o, and writing the equation ^-.-^^f.., 1 in 

P q r 
the abbreviated form pqr^ the equations of the sides of one tetrahedron are 

abCf a be, a be, a bo, 
while those of the inverse tetrahedron are 

abe, a be, a bo, a bo. 
The two systems of planes form an octohedron. 

I beg^n with the clearest case of intersection, in which the comers of the 
inner tetrahedron simply project through t^e rides of the outer one, without 
crossing its edges. 

Let the planes of the tetrahedra be represented by the following groups of 
planes, opposite planes (not necessarily parallel) written under one another, 

abe, abe, a be, a be, 

Imn, Imn, Imn, Imn. 
The secondary pyramid made by the plane Imn with the three planea 
abe, abe, abe has its four points (x^t/QZ^, &c. determined by the four 
systems of equations 



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48 



abef 


imn 

1 


l««) 


tmnl 


Zbe[ 


aici 


56«[ 


ibe\ 


abc) 


5Jc) 


ubV 


ale) 


which give the folkywing Tallies:^ 






«o - -a. 


y»- 


-6. 


«»--«'. 


9i - -a, 


fi- 


fcwi(a-D 


em»(a— n 


aln{b^m) 


9t' 


->, 


. _cft.(J-») 
"'miau+ctf 


_ flZm(c-ii) 




• (,am+bd' 




'*"j.(a«i + 6l)' 


#*- 


H' -e. 



' The other leoondary pyramids are obtained from the first bv changing simxil- 
taneonaly the signs or any two of a ft o and of the oorresponding pair of { m »• 
Now, changing the ngns o€ ablm and leaving the signs of en unchanged is 
equivalent to changing the sig^ of w and y and retuning those of t and of 
all the constants, and so with regard to the other two permutations. Hence 
it is dear that the sum of the sixteen x*s is identically equal to zero, and so 
of the ys and s's. It follows, then, that if these nzteen points be considered 
o the centres of spheres of equal weight, the centre of gravity is at the 
origin. If, therefore, the four secondai^ pyramids have all the same volume, 
the proposition will be proved. 

Conrnder the hexahedron (or parallelepiped) 

aOO, ObO, 00c, 
aOO, 06 0, 00c, 
the twelve diagonals of the foces of which are the edges of the two tetrahedra 
(a b c) and (a b c). Let tu suppose that its alternate oomen are cut off by 
the four planes Imii, lam, Imn, Imn, 

These planes will cut off equal lengths upon the corresponding edges of the hexa- 
hedron, and therefore the pyramids cut off will be of equal volume^ as may eanly 
be shown. Next, if we consider each plane of the type (/ m a) as the common 
base of two pyramids, one cut off from the hexahedron, and the other from 
the original tetrahedron (abe), we see that these have the same altitude, 
and that the proportion oetween their bases is the same at all the four 
comen. Hence the pyramids cut off from the tetrahedron are also all equal 
inter te. But this bein^ the case, it has already been shown that their centre 
of figure is at the origin. Hence the centre of figure of the solid about the 
origin, that is, of the remainder of the tetrahedron (a b c), is also at the 
oriffin. 

I have chosen the simplest case for demonstration. The general proposition 
may be inferred from the principle of the permanence of eouivalent formfl^ 
or, with a little more trouble in picking out the details, may oe proved inde- 
pendently. 

Cob. 1. — ^The solid formed about the origin by any number of such inter- 
secting tetrahedra, and by hexahedra of the same system, has also the origin 
for its centre of figure. 

Cob. 2. — Cleavage parallel to one of the coordinate planes simply shifts the 
centre of figure along the corresponding coordinate axis. 

Cob. 8.-— If the solid be made up of any number of tetrahedra and hexa- 
hedra about the same axes and having their oomera on the same four dia* 



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49 

gonals, then the effect of deavage in a tetrabedral plane is amply to shift the 
centre of figure along the diagonal corresponding to the cleavage plane. If 
there are to be no hollow spaces, there mnst of coarse be at most only one 
hexahedron, and one direct and one inverse tetrahedron. 

Octohedral cleavage is thus seen to be more complex than bexahedral, as 
might be expected. 



NoTB OTSf Question 2740: by Fbofesbob Catlby. 

The envelope of the curve 

A cos20+Bsin20 + Cco8 9 + Dsin0 + E»O, 
(where A, B, C, D, E are any fanctions of the coordinates, and is a variable 
parameter,) is obtained in the particular case E^O (Salmon's Higher Plane 
Curves, p. 116), and the same process is applicable in the general case where 
E is not ss 0. From the great variety of the problems which depend 
upon the determination of such an envelope, the result is an important one, 
and ought to befamiUarly knoum to students of analytical geometry. We 
have only to write 2 » cos + 1 sin 0, the trigonometrical functions are then 
given as mtional functions of z, and the equation is converted into a quartic 
equation in 2r ; the result is therefore obtained by equating to zero the dis- 
criminant of a quartic function. The equation, in fact, becomes 

that is, A(«* + l)-B»(«*-l) + C(«» + «)-Dt(»»-«) + 2Eaa = 0; 
or, multiplying by 12 to avoid fractions, this is 

(<i,J,c.<f.«)(«,l)*-0, 
where a->12(A-Bi), i»8(0-Dt), c»4E, 

« = 12(A + B0, d«8(C + Dt); 
and substituting in 

(ae-46ci + 3c*)»-27(actf-ad»-68e + 26cd-c»)2 -. 0, 
the equation divides by 1728, and the final result is 
{12(A2 + B2)-8(C2+D2) + 4E3]' 

- {27A ((?- D«) + 54BCD- [72 (A« + B*) + 9 (C? + D^)] E + 8E»] « - 0. 

It is to be noticed, that in developing the equation according to the powers 
of E, the terms in E^, E^ each disappear, so that the highest power is E^; 
the degree in the coordinates, or order of the curve, is on this account some- 
times lower than it wovHd otherwise have been. 



2823 (Proposed by Professor Sylvesteb.)— Show that on a chess-board 
the chance of a rook moving from one square to another without changing 
TOL. XI. P 



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50 

colour IB f; but that without altering^ the equality of the number of the 
black and white squares, bat only the manner of their dbtribntion, the 
chance may be made eqnal to J. 



I. Solution by Mosgan Jsinci^s, M.A. 

Wherever the rook may be placed on the board, there are in each of the 
two lateral directions in which the piece can move 3 squares of the same and 
4 of a different colour to that of the square on which the rook stands. The 
required chance is therefore f . 

No possible distribution of the colours can be made which shall raise the 
chance to i wherever the rook may be placed ; for if or be the number of 
black squares in a row, then oat of the 14 squares to which the rook may 
move 7 squares must be of the same and 7 of different colour to that of the 
square occupied by the rook. Hence, placing the rook on any of the x black 
squares, we should have x—1 other black squares in the same row, and there- 
fore 8— J* other black squares in the same column; and placing the rook on 
any of the S—x white squares, we should have x squares in the same row, 
and therefore 7—x black squares in the same column, whence follows the 

equation ' x (9-ar) + (8-ar) (7-ar) =82 (1). 

The only solution of this is rr— 4; which shows that there could be only 4 
black squares in any row, and in like manner in any column ; but this is 
inconsistent with the supposition in (1), that if there are ^ or 4 black squares 
in a row, there will be 9— j; or 5 in some of the columns. 

Hence the chance cannot be ^ for every position of the rook ; but by sup- 
posing that every position of the rook is equally probable, we may, by various 
distributions of the colours, make the total chance for all positions of the 
rook i. 

Let Xi,X29 ,... Xg be the number of black squares in the rows, 

yi, y2» • • • • ys » » »> columns ; 

then X1 + X2+..,, +xs^S2, yi-»-y2+ •• •• +^8 «« 32 (2,3); 

and also the following equation is easily obtainable, 

a^i(a?i-l) + arj(ar8-l)+....+(8-ar0(7-a?,)+.... > ^32x14, 

+yi (yi - 1) +ys (y2-i) +....+ (8-5^1) (7-y,) +....) 

which reduces to Xi^+ +yi^+.. .. = 16x18 (4). 

It is easily seen that, if Xi, d?2* • • • • i yi* ^3* • • • • ^ ^ ^^^ ^^ solutions, then 
8— *!, 8— d:2» . • • . f 8— yi, 8—^2, .... is another set, which may be called a 
complementary set. 

To facilitate the finding of solutions of (2), (3), and (4), I have made the 
cueumption (involving possibly a loss of generality) that 

xi^+x<?+ .... +a'8^ = 16x9 = y,«+y2'+.... +^8* (6,6). 

Supposing that there are a rows containing 1 black square, b rows containing 
2 black squares, and so on, we have 

a + 6 + c+...*=8, a + 26 + 3c+...«32, a + 45 + 9c + ...= 144 .... (7,8,9); 

whence we obtain 

6 + 3c + 6rf + 10cf ... = 66, c + 3<i + 6c+... =32 ....(10,11). 



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The following sets of values of Xi, X2, . . . . ^s ^^^^ ^ found, I think, to be 
all which can be obtained from the positive integral solutions of (7), (10), 

and (11). [6, 6, 4, 4, 4, 4, 2, 2] self-complementary 

[6, 6, 5, 4, 3, 3, 3, 2] [6, 5, 5, 5, 4, 3, 2, 2] 

[7, 5, 4, 4, 4, 3, 3, 2] [6, 5, 5, 4, 4, 4, 3, 1] 

[7, 5, 5, 3, 3, 3, 3, 3] [5, 5, 5, 5, 5, 3, 3, 1]. 

The following squares are specimens of distributions obtainable from these 
sets of values: — 



4 


w w 


W W 


B 


B 


B B 


4 


w w 


W W 


B 


B 


B B 


2 


w w 


W W 


B 


B 


W W 


2 


w w 


W W 


B 


B 


W W 


4 


B B 


B B 


W 


W 


W W 


4 


B B 


B B 


W 


W 


W W 


6 


B B 


B B 


B 


B 


W W 


6 


B B 


B B 


B 


B 


W W 



2 


w 


W 


W W W B W B 


4 


B 


W 


W W W B B B 


3 


B 


W 


W W W "W B B 


3 


B 


B 


B W W W W W 


3 


B 


B 


B W W W W W 


5 


B 


B 


B B B W W W 


6 


B 


B 


B B B W B W 


6 


B 


B 


B B B B W W 



44446622 



75533333 



[Professor Stlvksteb remarks that any anallagmatic distribution of the 
black and white squares will satisfy the condition. For, in such a square, 
whether the motion is from a row to a row or from a column to a column, 
there will by its definition be as many passages from like colour to like as 
firom a colour to its opposite. Suppose an anallagmatic square is made iso- 
chromatic, i. e., having always the same difference (positive or negative) be- 
tween the black and white colours in each row and column (see the diagram 
at the end of Vol. X. of the Reprint). Then, putting n for the number of 
squares in a side, and x,n—x for the number of the separate colours respec- 
tively, it is clear from the above solution that we must have 

2ar(»-*) = J2».^zl, .-. x{n-x) » !^Zf, i.e., x , 



n±^n 



which proves that an anallagmatic square cannot be made isochromatic unless 
the number of places it contains is a perfect fourth power. This disposition 
springs directly from an improved form given to Newton's complete rule 
for limiting the total number of imaginary roots in an algebraic equation.] 



II. Solution by Matthew Colldts, B.A. 

1. As each line and column of a chessboard contains 4 white and 4 black 
squares, therefore upon whatever square a rook be placed there are 3 other 
squares of the same colour, and 7 other squares altogether in the line or 
column in which the rook can move, and hence his chance of moving into a 
square of the same colour as that be left is plainly « f . 

2. If the Une bisecting two opposite sides of the chessboard be the boundary 
between the white and the black squares, it is plain the square common to 
any line and column has, upon that line and column taken jointly, 10 other 
squares of the same colour as itself; and as there are 14 other squares alto- 
gether upon that line and column, hence in this case the chance of a rook 
moving into a square of the same colour as that it left is » -^ = f . 



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52 

8. If the diagonal dividea the white from the black aqoares, and if the 
aqnares through which it paaaes be white and black alternately, the required 
chance of a rook not changing colour when moved would plainly be 

14.7 + 10,9 + 6.11+2.13 _ 6 
82.14 8* 

4. But if the four white squares on the diagonal occupy half of it, the 
other half running through four black squares, then the required chance is 

•^ (4.6 + 6.7 + 6.8 + 5.9 + 5.10+8.11+2.12 + 1.13) r=???. 



82 . 14 ^ '448 

6. Lastly, if one row consist of eight white squares, and another row of 
eight black squares, the remaining six rows of the chess-board being left 
unchanged, the required chance is then plainly 

8.10+24.6 _1 
82.14 "2 



274L (Proposed by S. Tebat, B.A.) — A moveable event (depcpduig on 
the moon) happened in the year y ; show that it will happen again in the 
year ^ + 19(28^— 5a +86) ; where a may be 0,1,2,3; b the number of 
completed centuries since the year y which are not leap years; and t any 
arbitrary integer. 

Solution hy ike PaoFOSEB. 

Let T B f^ + 4»-f-a be the year ; then the number of days in tins interval 

is 865(4n + a) + fi-A«7(208» + 62a) + 5fi + a-6; 

therefore Sn + a— h must be diviMble by 7 ; 

therefore ii-»7m— 3a+d&, and 4it + a » 28m— lla+125, 

which must be divisible by 19 ; 

therefore m » 19^-3a + 5i, .'. 4fi+a » 19 (28<-5a + 86), 

and Y = v + 19(28/-5a + 8ft). 

JExan^le : Easter-day cannot happen later than April 25th ; the last time 
this took place was in the year 1734 ; when will it happen again? Let t=^0, 
assO, isl; therefore T — 1886, the year required. 



2598. riVoposed by G. M. Suith, BJ^.)~A uniform rod (mass » /u) is 
placed inside a spherical shell (mass » m, radius = a) which, under the in- 
fluence of g^vity, rolls down the sur&ce of a rough sphere (radius «r) ; find 
equations for the movements of the rod and shelly there b&ng no fiiction 



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53 

between them; and hence show that, if a particle fi is placed near the lowest 
point of a spherical shell (m) which performs small oscillations on a rough 

horizontal plane, the length of the simple equivalent pendulum is — ^ a. 



Solution by the Fbofossb. 

Let the horizontal and vertical lines through the centre O of the sphere be 
taken as axes of x and y respectively ; let (x, j/) be the coordinates of the 
centre C of the shell ; ^ » angle through which the shell has turned ; ^ « 
angle between radius and Ojf; wm COy; c » perpendicular from C on rod. 
Then 

« — (m4-/u)^8y + /ic^cos^S^. 

Substituting for x and y their values in terms of $, and selecting the co- 
efficients of 80 and 8^, we obtain, by equating these coefficients separately to 
zero, the equations 

(2«+M)(a+0S-^[008(*-<')g--(*-»)(g)'} 

+ (m+ii)gcoae — 0, 

c(a+r){co8 (♦-») g + sin(^-e) gyj-(*»+o»)g +gc^4> - 0. 

In the particular case in which the radius becomes a partidei and the sphere 
a horizontal plane, these equations become 

ac rin 4. ^ - (i»+ e») ^ + cj sin ^ - 0, 
dP fir 

which give (by putting ^ — ^ir— iS, where jS is small) 

showing that the length of the simple pendulum » a. 

In the case in which the sphere reduces to a pkne of inclination » t, our 
equations become 

(2»+M) « ^ + /«: ^£?i^ + (»+m) J Bin .• - 0. 
«,rin(^+.-)^ -(*»+c»)^ + ^coe ♦ = 0. 



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54 

222L (Proposed by W. S. Buskside, M.A.) — 1. Show how to deter- 
mine the locns of the feet of perpendiculars from a fixed point on the gene- 
rating lines of a system of confocal qaadrics. 

2. Prove that this locns is described by foci of the plane sections passing 
through the fixed point. 

Solution hy the Pboposeb. 

1. Since the foot of a perpendicalar on a generator is the foot of a per- 
pendicular on a tangent plane through that generator, the question ( 1) is 
reduced to first finding the locus of feet ot perpendiculars on tangent planes 
to one surface of the system, and then eliminating the arbitrary parameter 
which enters by means of the equation of the surface itself. 

2. Consider the section of a surface of the system by a plane P, and let 
one of the foci of the section be at F ; this being so^ reciprocating origin at 
F, the section of the reciprocal surface by the plane P is a circle ; con- 
sequently the section of the asymptotic cone of this burface is a circle ; but the 
asymptotic cone is similar and similarly placed to the cone generated by lines 
passing through F perpendicular to the tangent planes through F to the 
given surface ; therefore one of the focal lines of this cone is the perpen- 
dicular to the plane P at F. Finally the focal lines of the tangent cone are 
the generators of the confocal hyperboloid of one sheet through the vertex. 



2795. (Proposed by S. Tebat, B.A.)— Find the average square of the 
distance between the centres of the inscribed and circumscribed circles of a 
triangle inscribed in a given circle. 



I. Solution hy the Bev. J. Wolstenholice, M. A. 

If a be the radius of the given circle, * radius 
of an inscribed circle, square of distance between 
the centres is a^ — 2aa;, and the question is 
equivalent to finding the average value of x. 
Now, considering one point A of the triangle 
fixed, which is clearly allowable ; AO the tan- 
gent at A ; P, Q the other angular points of the 
triangle, 

zOAP«a, Z0AQ = 4>, (4>>a), ar = 4a8inie8ini(<^-e)cosi<^, 

/*' /'"^4fflni«8m^(^-0)co8^<?e<2^ 
.*. avei'age vahie of a; » a 







d0d(t> 



(See Book of Mathematical Problems, Ex. 1036); 

therefore average square of distance between the centres is Sa^ M — _ J. 



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65 



II. Solution hy the Pbofosbb. 
Let D be the distance between their centres ; then 

=^ R2 (1-8 sin ^A sin JB sin ^C) 
«R2-4R2siniA {sin^ (A + 2B)-sin ^A}. 

Supposing the vertical angle to remain constant while B varies from to 
ir— A, we find the sum of the squares of the distances 

= 2 /*' f*'^ D2<2AdB « 3R2(ir«-8). 

o o 

The number of triangles — 2 /*' j dk d^ = ir^. 
o o 

Hence the average square of the distance » 3R^ / 1 j. 



2797. (Proposed by the Rev. J. Wolstenholmb, M.A.)— In any com- 
plete quadrilateral, the radical axis of the three circles whose diameters are 
the diagonals will pass through the centres of perpendiculars of the four 
triangles formed by the four straight lines. 



Solution hy Matthew Collots, B.A. 

Let ABCD (Fig. 1) be the quadri- 
lateral; then the circles whose dia- 
meters are the three diagonals AC, 
BD, £F will have their centres at the 
middle points G, H, K of these dia- 
gonals ; the three altitudes of the tri- 
angle ABP, viz., AOA'. BOB', FOP', 
meet in one point O (its ortho-centre) ; 
therefore G^, perpendicular to AOA', 
is parallel to BC and bisects AA' in g\ 
similarly, HA perpendicular to BB', 
and Kk perpendicular to FF^, bisect 
BB' and FF' in h and *. 




Fig. 1. 



Now, the square of the tangent drawn from O to the circle (G) 
= 0G2- AG^ - Og^-Xf = OA . OA' ; similarly, the square of the tangent 
from O to the circle (H) = OH^-BH^ - Oh^-Bh^ - OB . OB', and the 
square of the tangent from O to the circle (K) = OK^-FK^ = 0;fc«-F*« 
-OF. OF'. 

But OA.OA':*OB.OB'«OF.OF', therefore the tangents from O to 
the three circles are equal to each other, consequently the ortho-centre O of 
the triangle ABF is a point on the radical axis ot these three circles. For 
like reasons, the ortho-centres of the other thr^ triangles formed by three 



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other ddes of the quadrilateral ABCD (yu., the triangles ADE, BCE, CDF) 
He also upon the same radical axis. 

CoBOLLABT.— Ag the three circles (G), (H), (K) are proved coaxal, their 
centres, viz., the middle points of the three diagonals of the complete quadri- 
lateral ABCDEF, lie in one straight line. A rimple proof of this theorem 
may also he ohtiuned as follows : — 

Let ABCDEF (Fig. 2) he the com- 
plete quadrilateral; AC, BD, £F its 
three diagonals. Through B,C,D,E,F 
draw lines parallel to the sides of the 
angle A, forming the parallelograms 
ABGD, AEHF, &c 

Then (hy Euclid I. 43), parallelo- 
gram CK»AC»CK', nnd therefore 
parallelogram LE»L'E'; hence (hy 
Euclid I. 48, ex absurdo) the points 
C, G, fl lie in one straight line. 

Now, as the diagonals of a parallelogram hisect each other, the middle 
point of the diagonal BD is the same as the middle point of AG ; and, for a 
like reason, the middle point of the third diagonal EF is the same as the 
middle point of AH. But the middle points of AC, AG, AH lie plainly in 
one straight line parallel to CGli ; therefore the middle points of the three 
diagonals AC, BD, EF of the complete quadrilateral ABCDEF lie in one 
strught line. 

Note. — Several new and curious theorems can he easily ohtained from the 
foregoing theorem and its corollary hy means of the methods of reciprocation. 
and inversion. 



I ,1.0-^1 


•A 


<^h 


I' 


/ 


U 


-"// 


V 


r 




Fig. 2. 


H 



(Proposed hy A. W. Pauton, B.A.) — ^The equation connecting 
the distances (ri, r^ r^ of any point on a Cart^an oval from the foci is 

ifi-i) o*ri+ (7-a) i8*r2 + (a-jS) -^r^ = 0, 
where a,fi,y are the distances of the foci from the triple focus. 



Solution hy W. S. McCay, B.A. 

The Cartesian oval may he written 

Zri + wrj+wrs— 0, or Xfi + fira « d (1.2). 

Now to reduce the first of these to the second hy means of the identity 

An2+Br2«+Crj3=:D (3), 

(where A, B, C are the distances hetween the foci, and D =:: ABC) we get 
hy suhstituting r^ in (3) from (1) 



Ar^^-^ Bra« + C (^I^V- D. 



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That this may be identical with (2), we must have 



P . m» 



An»+CP 



.(6,6). 
...(7). 



ABC B»?+Cj»* 

A5 B* 

The last eqnation becomes by means of (5) \^ I /i^ ^ — : — 
Now since (2) may be written 

(.v..-.V-^5V^'(n.-..,^)-o. 

it follows that if (a, jS, y) are the distances of the foci from the triple focus 
^centre of circle xVi"-/*^ r^^ « ^4)' ^'«-m'^ * 0. 

Hence (7) becomes ? : i» « Aa* : B/B* ; 

so too m : i» = B/S* : Oy*; 

therefore llmln^ (fi-y)^ : (y-a)fi^ : (a-0)y^. 



2670. (Proposed by C. T. Hudson, LL.D.)— An observer, seated in the 
aisle of a cathedral and looking westward, sees the horizontal lines above the 
arches converging to a point before him, and consequently cutting the horizon 
at a certain angle. On his facing northwards, he sees that the same lines are 
now parallel to the horizon. Required the curve that they will appear to lie 
in, as he gradually turns his head through 90^. 



SoluHon by tXe Pboposes. 

This problem is indeterminate ; for the plane on which the given lines are 
seen projected at any instant, will at that instant be referred to various 
distances according to the fancy of each observer. 

(i.) Let this plane be supposed to be vertical, and always at a fixed distance 
from the observer; then the ultimate intersections of its various positions 
will be a vertical cylinder of which the axis passes through the observer, and 
a plane through bis eye and the given horizontal lines will cut the cylinder 
in an ellipse, which therefore is the required curve. 

(ii.) Let that portion of the given horizontal line which is seen distinctly 
at any instant be supposed to be referred to its real distance from the ob- 
server at that instant. 

Let O be the observer, AB tlie given horizontal 
line, Oy and Ox lines perpendicular and parallel to 
AB in the plane drawn through AB and O, QPthe 
portion of AB which is seen distinctly when the 
observer is looking along OP, but which is seen 
projected into the position PR at right angles to 
UP. Let also AO^^, POa?=0 ; then the equation 
to PR is scoaB + ysinB ^ccoaecO; 




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and the nltunfttemtenectioiis of all its poeitioiis obtained by varying lie in 

the cnnre a* =» 4c (c— y) ........* (1). 

The sensation of the enrvatnre of Afi is, according to this hypothesis, pro- 
duced by the attention being fixed on the extremity of the line BP, which 
always moves so aa to to«cfa the parabola (1), and at the same time remain at 
right angles to the visual ray. 



2762. (Proposed by the Esitob.) — Find the envelope of an ellipse which 
has one vertex in the curve of a given parabola, and touches at its adfaoent 
vertex the extremity of a fixed double-ocdinate of the parabola. 



I. SoImHoh hy J. J. WxLKBB, M.A. 

Taking the ordinate of the parabola and the axis of 
the ellipse which passes through the vertex of contact 
as axes of y and x, the equation to the parabola will be 
(n being the fixed ordinate) 

ya-2ny ■¥px « (1), 

and that of the ellipse 

y' +-^!^ + -?PiL«0....(2), 

where y is the ordinate of the parabola passing through 
the first vertex of the ellipse, since the axes of the 



ellipse will plainly be ^ + » and 



Equation 




(2), when multiplied by (]/''—»')' and arranged by 
powers of y*, becomes 

(j/2 + 2par)y'2_2«yV+ nV+P'-^^-Zpna* « 0, 
the discriminant of which, with respect to y* as variable, is 

(y^ + 2px) («^2 +pia^^^n^x) — nV» or i^a?^ (v^ + 2px—4,n^. 
Hence the required envdope is the parabola 

y^ + 2px—4!h^=^ 0, 

which has as ordinate the fixed double ordinate, and a parameter double 
that of the given parabola. 



II. Solution ly the Ret. J. Wolst£NHOLME, M.A.; B. Bills,* and others. 

The equation of the parabola being y- = 4aa:, let the coordinates of the 
fixed vertex be <ia?, 2aa, and of the moving vertex aA-, 2a\ ; then the equa- 
tion of the ellipse is i^Zf^)!_ + (y-2^«)' ^ i 
or 4 (x- «X3)2 + (o + \y (y - 2aaf « 4a2 (a^- \^)^, 



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or 4(ar-aa5)(ar + ai^-2aX*) + (a + A)*(y-2aa)«=: 0; 

and the enTelope is 

a2(y-2aa)< = {(y-2aa)«-8a(ar-ai^)}{<^(y-.2aa)2 + 4(a>-aV)], 
or (x-a«^»[(y-2ao)«-8a(ar+ao=)} « 0. 

The required eavelope is thtis seen to consist of the fixed doable ordinate, 
and a parabola similarly sitnated to the given one, of twice the dimensions, 
and having its vertex in the diameter through the given vertex. For the 
contact to be real, the vertices must be on opposite sides of the axis of the 
parabola. 



2800« (Proposed by J. Wilson.) — Three circles passing through a point 
P form a circular triangle ABC, and each side of this circular triangle or its 
continuation is cut orthogonally at the point P' by a circle passing through 
P; prove that the three circles described about the triangles PAP, PBP", 
PCF" are co-axal. 



Solution hy Abcheb Staitlit. 

By inverting the figure with respect to the point P, it will really be 
found that the theorem is the inverse of the well-known one, that the per- 
pendiculars are concurrent which are drawn from the vertices of a triangle to 
the respectively opposite sides. 



2688. (Proposed by J. Gbiffiths, M.A.)— A variable triangle circum- 
scribes an equilateral hyperbola, and is such that its wine-poiwi circle passes 
through the centre of this curve ; prove that the locus of the centre of its 
eircwnacribing circle is the hyperbola in question. 



Solution hy the Rev. J. Wolbtbnholhi, M.A. 

If an equilateral hyperbola be inscribed in a triangle, the centre of the 
hyperbola lies on the polar circle of the triangle : if. then, the centre lies on 
the nine-point circle, it must lie on the circumscribed circle, since the three 
drcles are coaxal. Let O be this centre, ABC the triangle, and let the 
asymptotes of the hyperbola meet the circumscribed circle in P, Q. Then, 
since OPQ, ABC are triangles in the same conic (circle), their sides will 
touch one conic ; but five of them touch the hyperbola, therefore also the 
sixth (PQ) touches the hyperbola: it will be bisected in the point of contact, 
which will therefore be the centre of the circumscribed circle of the triangle* 
Hence, if a rectangular hyperbola touch the sides of a triangle, and its centre 
lie on the nine-point circle of the triangle, the centre of the circumscribed 
circle lies on the hyperbola; which proves the theorem. 



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2654. (Proposed by W. S. Bubkbidb, M.A.) — Determine the polar 
reciprocal of the quartic curve 

^ + «V+a^3+2iiy«(Lp+my-HftB) « 0. 



Wnte J. 1. I 



SoUaion by the Pboposkb. 
for X, y, iB, and the qnartic 



18 transformed to 



X>+T«+Z2+2nrZf2«iZX + 2«XY = 0, and- the polar of a pomt ajSy 
becomea aTZ + jBZX + 7XT ^ 0. So the qnestion is rednced to finding 
when these two conies touch, which is easily done. 



2726. (Proposed by the Rev. J. Wolstenholkb, M.A.) — If a conic 
touch the ndes of a triangle and pass through the centre of the circumscribed 
circle, the director circle of the conic will touch the drcumscribed circle of 
the triangle. 

Solution by W. S. MoCat, B.A. 

If the conic be (te)* + (wty)* + (n»)* — 0, it follows 
from the general equation of the director circle given 
at p. 338 of Salmon's Conies ^ that the radical axis of 
director circle and circumscribing circle of the triangle 
is icot A.a? + flicot B.y + ncot C.» i» 0; 
and if this touch the circumscribing circle {ayz + hxt 
+ cxy — 0), we must have 

Pcoe? A + »i»cos» B +n«co8« C— 2m«ri^ cosB cos C 
- 2Pn^coB B-2m«i«cos2 C - 0, 

or (I cos A)* + («» cos B)* + (» cos C)* =■ 0. 

But this is the condition that centre of drcumscribmg drde (cos A, oqb B, 
008 C) should lie on the conic. 

[Mr. Wolstenholmb's proof is as follows :— 

Let O be the centre of the circumscribed circle, and let a tangent to the 
conic at O meet the circle in B', C; then the tangents from B^ (^ to the 
conic will meet in a point (A') which will lie on the circle, since ABC, AfWQf 
are triangles circumscribing the same conic These tangents being at right 
angles. A' is a point on the director circle; and since WCf is bisected in O, 
OA' passes through the centre of the conic, that is, through the centre of 
the director drcle. The <drcles therefore touch at A'.] 




2770. (Proposed by the Rev. J. Wolbtenholme, M.A.)— If a conic be 
drawn through S, S', the foci (both real or both impossible) of a given conic, 
and osculate the conic at a point P, the tangents at S, S' will intersect in the 
centre of curvature at P. 



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61 

SohOion by tie Fbofossb. 

The equation of any conic touching the g^ven conic ^ + IL b 1 at the 
point (a cos 9, ( sin 9) is 

if this osculate at the point $, L cosi^ 9 + M sin* 9 a> K ; 

if it also pass through the two fbd on the axis of x, 

L+N«0, A-5:!Vl + Loo8«9)-l-N; 

Also the pole of SS' is given hy the equations 

0+Lc(rfa)f+°°''=°'*(L 
a o 

But, from the former equations^ 



2(l+Lcoe»9)*+??L?i5^(L + M)y-0, M±?5r8in9+2a-N) « 0. 
a o o 



-L-N=. ** - ^«»°*^ 



and 



a3fflnS9+^oosS9 l + cu8S9* 

M + N 2A« ,^_^ a»-ft» , 

1-N (a«-^8in<fl 

L + M 252co8«9 



therefore ?-^^ ^ . - ^ . .^ therefore y =- iJZf: rin» 9 ; 

*9^ "^ 6 



1 + LC0829 a2sin<9* 



therefore f - *i?!!;! . ^ sin' 9 . ^ cos^ 9 ; 

a a'sin'9 o a' 

or the tangents at S» S' intersect in the point (^ " cos* 9, ""^ mn* 9], 
the centre of curvature at P. 

It may be noticed that, in any conic passing through SS' and touching at 
P, the tangent and normal at P, form with PS» PS' a harmonic pendl ; 
therefore SS' and the normal at P are conjugates with repect to this conic, 
or the tangents at S, S' intersect on the normal at P. 



2799. (Proposed by A. Mabtin.)— If n dice are thrown, what is the 
chimce of an odd number of aces turning up ? 



Soluiion hy the Rby. J. Wolstbnholkb, M.A. 

The chance is obviously . If the dice be regular tetrahedra, the 

chance is —> 

2 2""*"* 



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2789. (Proposed by W. S. M*Cay, B.A.)— If two drdes pass through 
the foci of an ellipse and touch the same variable tangent to the curve, the 

angle at which they intersect is constant and equal to 2 tan~i -. 




Solution hy the Rev. J. Wolbtekholmb, M.A. 

Let SS' be foci of an ellipse, TPT a 
tangent meeting the tangents at BB', the 
ends of the minor axis in TV, and the 
major axis in U. Then ZUTS'« Z BTS 
= I TSS'; therefore triangles TITS', UST 
are similar, and UT : US' = US : UT, 
or US.US'-UT2; therefore the circle 
through STS' will touch the tangent at T, 

so the circle through ST'S' will touch the tangent at T^; and the angle at 
which these circles cut ift S, S' is the sum of the angles STS', ST'S', and 
therefore the difference of the angles TS'T', TST'. 
But Z TSr- i z BSB', and Z TS'r = J external angle BS'B'= w- |BSB'; 

therefore angle at which the circles cut « ir— BSB^= 2 tan-i£. 

The Brst part of this problem I have already set in the form, " If a circle 
pass through the foci of an ellipse, and common tangents be drawn to the 
ellipse and circle, their points of contact with the circle will lie on the tan- 
gents to the ellipse at the ends of the minor axis." 



2780. (Proposed by Professor HiBST.) — The envelope of the chord com- 
mon to an ellipse and its circle of curvature is a curve of the fourth class 
which has three double tangents, one at infinity and the two others coinci- 
dent with the conjugate diameters equally inclined to the axis. Prove this, 
examine the curve, and consider the cases of the hyperbola and parabola. 



Solution ly J. J. Waleeb, M.A. 

The equation to the ellipse being b^x^-^cfiy^ = a%'^, if the coordinates of 
any point on it be a cos 0, b sin 9, the reciprocals of the intercepts made on 
the axes by the chord common to the ellipse and the circle of curvature at 

this point will be a = -5?L — , fi « -Zfl!l__, and eliminating between 
^ acos2e hem 29 ^ 

these equations, a^a^ + b^ff^ = (a^a^—b^fi'^^ is the envelope of the chord, — a 
curve of the fourth class, having as double tangents the lines joining the 
points aa±bfi = with the centre, — i, e,, the equal conjugate diameters; 
also the line a=0, jB=0. 

For the hyperbola, the sign of 6^ is to be changed, and the first pair of 
double tangents become imaginary. 

In the case of the parabola y^ =■ 4mxy the tangential equation to the 
envelope of the chord common to it and its circle of curvature at any point 



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is nmilarly found to be « » Zmfi^, a parabola having the same vertex with 
the given parabola, and as its axis the production of axis of the other. Its 
parameter is three times that of the given parabola. 

The equations to the above envelopes in Cartesian coordinates are 

(AV + aV-4a262)3+27(Z^2j^-aV) = and f/^ + l2mx^0 

respectively, the equations to the ellipse and parabola being b^a^ + a^y^ = a%'^ 
and y2 «■ 4mx respectively. From the former it -appears that the points of 
contact of the equi-conjugate diameters lie on a similar and coaxal ellipse, 
and that there are cusps at these points. 

[The equation of the envelope has been given by Mr. Wolstenholme, in 
his Solution of Question 2579 (2), (Reprint, Vol. X. p. 91) in the neat form 



(M)'^(M)*=^' 



which has cusps at the points 5- s= ^ » 2, and of course two at infinity. 

The form of the equation to the envelope given by Mr. Waleeb in Question 
2834 may be readily deduced from Mr. Wolstenholme's.] 



To express the distance between the centres of the ci/rcumscrihed and 

inscribed circles in terms of the radii of those circles. 

By J. Walmsley, B.A. 

Let O', O be centres of circumscribed and inscribed 
circles respectively of the triangle ABC. Join O'O, 
OB, OC. Produce BO to D, and join DO'. Di-aw 
O'P pei*pendicular to BD, and therefore bisecting it. 



Tlien 



B0 = . 



and 



therefore 

Hence 
therefore 



sin-^B 

DCO = DCA + ACO = DBA + AGO 
- OBC + OCB = DOC 5 

asin^B ^ 




2B$ia^B. 



DO = DC=^ . ^ 
Bin A 

DP2-0t»* ^ DO . OB - 2Rr. 



0'D»-0'02 



0'0« « R2-2Rr. 



In an analogous mannei^ if Oi be eentre and ti radius of escribed circle 
touching BO, we may show that 0'0i2 = R2 + 2Rri. 



2776. (Proposed by W. K. Clitfoed, B.A.) — Through A, the double 
point of a circular cubic, draw AB perpendicular to the asymptote ; if chords 
be drawn to the curve subtending a right angle at the double point, show 
that there is a fixed point in A B at which also they subtend a right angle. 



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Solmihn hf the RxT. J. Wolbtevhoucx, M.A. 

The equation may be taken y^^a? ^^^i the origin being 

o— a? 

the double point. If (I'lyi), (jts^i) be the ends of two chords 
AP, AQ at right angles to each other, XiX^-^-yi^^ » 0; and 

h—xi 0— a?, 

therefore (a?i + a) {x^ + a) « (6 — tj) (6 — a?j), therefore «i + a^ = b — a. 

Bat the equation of the circle on PQ is 
(a?-ar,)(ar-a!^ + Of-yO(y-ya)=0, or a»+i^-a?(jri+«i)-y(yi+ya) - 0; 
therefore, when y =» 0, « » or a? «> a?i + x, » 6— a, 
giving a fixed point on AB at which PQ subtends a right angle. 




2544. (Proposed by Professor Ball.)— There are two values oid for 
which a, ^, 7, the roots of the cubic a** + Zlv^ + 3cj: + d — fulfil the linear 
relation Aa + BjB + C7 •- 0. Show how to find the quadratic equation of 
whichlthese values of d are the roots. 



I. Solution htf the Bby. J. Wolstxkholmb, M.A. 

The relations a + i3 + 7- -_, Aa+Bi3 + C7 = 0, fiy-^ya+afi ^^ 
a a 

Icad^at once to a quadratic equation in 7 ; whence 

86 A«+B2-C(A + B) _o 

^*'*"^«""TA2 + B* + C»-BC-CA-AB=^ 
8ac(A-B)3+ 9ft-AB-_ 
'^»'^»'*aHA»+...-BC-...;-"- 
But -(c?i + <y= +a(7i8+73»)+3ft(7i2+72S) + 8c(7i + 72), 

and didi = 7172 (071' + Sbyi + 3c) (0732 + 3672 + 3c), 

which can be immediately expressed in terms of m and n. 



II. Solution hy J. J. Waleeb, M.A. 
By the terms of the question, 

BjB + C7 = -Aa (1), afi-^ay ^~-aa-Zh (2), 

a^7 + a (i8 + 7) a « 3c (3). 

From (1) and (2), a(B-C)i8 - a(C-A)a + 36C (4), 

a(C-B)7«a(U~A)a + 36B (5); 



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-a5(B-C)»jS7 - o»(A-B)(A-C)a» + 3aA{2BC-A(B + C)}a + 95»BC 

(6), 

Eliminating /3 and 7 from (2), (3), (6), there results 

a»[(A-B)(A-C)-(B-C)«}a»+3a5{B« + C«-A(B + C)}a 

+ 3[ac(B-C)« + 362BC]«0 (7); 

also aa» + 3*a' + 3ca + (J « (8). 

The resultant of (7) and (8) is 

a'^d^ + 9a' { (3^»'2- 2aV) (a'J - aJ') - a'U {a'c- ac') } d 

+ 81c''(a'J-.aft')H27c'(a'c-.aO{a'(a'c-aO-3J'(a'6-a4';} -0... (9), 

if we write a'= aa{(A-B)^A-C) + (B-C)»}, 

&'-=a6{B2+C2-A(BfC)}, 
c'=ac(B-C) + 362BC, 
whence o'6-ay=a34(A2+BC), a'c -oc'- a {ac (A- B) (A- C)-362BC}. 

Making these substitutions in (9), it becomes divisible by a\ and is the re- 
quired quadratic for tf. The coefficients are not capable of any important 
rodnction, as far as I have examined them. 



1587. (Proposed by the Eev. T. P. Kiekman, F.R.S.)— Apollo and the 
Muses accepted the challenge of Jove, to vary the arrangement of themselves 
on their fixed and burnished conches at his evening banquets, till every 
three of them should have occupied, once and once only, every three of the 
couches, in every and any order. In how many days, and how many ways, 
did they accomplish the feat, keeping one arrangement of themselves through 
all the solutions P Required two or more ot these solutions, clearly indif* 
Gated, so as to save space, by cyclical operations. 



Soluiiafi hy the Pbopobbb. 

The solution is the positive half of the group of 10. 9 . 8 . 2 made with 
10 elements, which was first given by me in the Memoirs of the Literary 
and PhilosopMcal Society of Manchester^ 1861, and whose title I gave thus 
in their Proceedings five or six years ago : — 

10.9.8.2 =» 1 + 180q2 + 14458 + 452418 + 80381 + 2704,1, 

+ 802,14 + 18081, + 144io + 90^ +240ggi + 363, : Q - 2520. 

The positive half is the first six terms on the right, comprising, besides unity, 
the natural order, 180 substitutions having each a circle of 8 and a circle of 
2, 144 substitutions having each two circles of 5, 45 having each four 

TOL. XI. H 



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circles of 2, and 2 elements undisturbed, 80 having three circles of 3 and one 

undisturbed element, &c. 

This group of 10 . 9 . 8 . 2, ns well as every group so given by its title, can 

be easily constructed from its title ; but it is impossible to state the rules 

here. It is enough to give one of the 2520 equivalent groups of 10 . 9 . 8, 

thus, 

1234567890 1234567890 1234567890 
+0324897561 3215648970 1473692580 
+3017594862 3126459780 1327984650 
+2107864593 4567891230 1742853960 
+6280715394 5648972310 1685739240 
+3814069275 6459783120 1956278430 
+ 8243907156 7891234560 1869425370 
+8639520147 8972315640 1598346720 
+ 4261539078 9783126450 
+ 2165437809 

under the form (A);G;H where G;H is a group of 9.8, the product of 
two groups G and H. (A) is no group. The 10 . 9 . 8 cannot be written as 
the product of so few as three groups. We have the g^oup by adding 
to G;H its nine derivatives by the substitutions of (A). If these be 
1, Ai« A2,. . . . A9; ly Sfh Sf2' " * Sfs those of G ; and 1^ ^. . . . A7 those of H ; 
the group 

(A);G;H = G;H + Ai;G;H + Aj;G;H+ .. .. + A9;G5H, 
will be found to stand the test, that every product of its substitutions, if 
1 =■ Aq = ^0 = ^o» is of the form A ; ^ ; h^. Neither the algebraic nor the 
tactical demonstration of this can find space here. The group proves itself. 
These 10 . 9 . 8 arrangements are one of the 2520 solutions of the problem 
which Apollo's party dined out day b}"^ day for nearly 5000 years, at some 
expense of Jove's nectar. There are three, and only three, solutions which 
have in common the substitutions of G, (A);G;H, (A');G;H', and 
(A");G;H". No two solutions have more than eight arrangements in 
common besides unity. The above group (A);G;H is necessary as to the 
number, and sufficient as to the form, of its arrangements : necessary, because 
Apollo, Clio, and Urania cannot seat themselves in every way in fewer than 
10 . 9 . 8 arrangements ; and sufficient, because if in e and *, any pair of 
arrangements E, L, and M were seated alike, the group would contain the 

substitution — , which has KLM undisturbed; but the positive title has 

no substitution showing three elements undisturbed. From (A);G;H 
once written out can be formed all other solutions, by the operation 
6;(A);G;U;0'^, where $ is any substitution not before used. Thus, 
e ^e-' may be 2134567890 for a second solution. 

Note. — The unfashionable reader who wishes to learn how to find all possible 
groups made with n elements, to determine their titles and the number of 
their equivalents, and from the titles to construct the groups by a direct 
tactical method, without troubling himself with congruences, has only to 
fish up, if he can, from a basket in Manchester^ a Memoir of mine which has 
lain there for some years. Our learned societies agree with me in the 
opinion that it is absurd to waste English paper on sudi upstart subjects as 
Groups and Polyedra, which are fit for nothing but make-believe prize- 
questions of the French Academy. They have done quite enough in phicing 
it beyond dispute in their Proceedings that I have thoroughly discussed 
both these vast theories, as difficult as they are unvalued. 



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2847. (Propoied by M. W. Ceoftok, F.R.S.)— Prove that 

c*D\ «-*«• - (l + 4A*)-*€"i"^, where D^~ 

dx 



SolnOian by J. J. Walkbb, MJl. 
Expanding the symbolical operator, 



^D..,-*^^n.*I>'.*^. 



-te" 



But by the known formula for the saccesdve differential ooeffidents of c 
with respect to x, 

€*^ A D«. €-*** = « ^ A* + A*2 (2x)\ 

(1.2.3)» (1.2)2.3 ^ ' 

^ ^A»*»(2^)4+ *!^(2*)«, 



1.2.3 " ' 1.2.3' 



It will be found on examination that the rth term (or line) of the ^h column 
in the sum of the above developments is equal to 

^ ^ 1.2...(j,-l) 

^ (2r + 2p-4).(2r-f-2p-5)...(2p-l).(4Alr)'"^ 

1.2...(r-.l).r...(r+p-2).p.Cp + l)...(r-l).(2)^-"' 

when r is greater than p ; but when r is equal to or less than p, the denomi- 
nator of the last factor is 1 . 2 ... (r— 1) . p . (ji + 1) ... (p + r - 2) . 2^"*. In both 
> this factor is equal to (^^" ^) (2p + D - - (2p + 2r- 6) (4^;t,)r-i ^^^.^ 
1.2...(r-l).2*-* 

multiplied by (— l)*"'^ is the rth term in the development of (1 + 4Ak) ' . 

From this it results that f*^. €*»". €"**■= (l + 4Wr)-*c^****j and multiplying 
both sides of this by c" , there results for c . c" the value given in the 



[This theorem, which the Proposer has found to possess important appli- 
cations in the Theory of Errors of Observation, seems deducible in the most 
direct, though not the most elementary, manner, by putting ^'^^ for ^(x) 
in Poisson'b famous transformation {TraiU de Micanique, Tom. II., p. 356)^ 
which gives 



^{r)^l-f €-V(* + 2A*«)rf«.] 






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(Proposed by W. S. McCat, B.A,)— If a conic pasg throagb the 
foar points of contact of tangents to a cabic from a point (A) on the curve, 
and throa$?h two other points (B, C) on the cubic ; then A is the pole of BC 
with regard to the conic 

Solution by F. D. Thokbok, M.A. 

The equation to any cubic may be 
written 

S=jTV+*yU = (1), 

where U, V are conies intersecting in 
four points A, B, C, D on the curve, and 
a and y are straight lines intersecting 
on the curve, in the point O suppose. 

The equation to the polar conic of (1) 
with reference to the point (a?', y', z") is 

/ dS J dS / dS ^ 

dx dy dz 

which may be written 

a?^ + VU + ^Pv+*^P|* = ^ (2)» 

where P^ denotes the polar of (j:',y, z') with respect to U. But if (ar', y', «') 

be on the line OP which touches U, a/ss 0, and P becomes a?-^. 

Hence the equation (2) reduces to 

Ar/U + ar(p^+^g,)-0. 

Therefore, if U is itself the polar conic of the point P, 

P +>ty^=0 identically. 
dor 

Therefore P^ coincides with y — 0, ».e., with EP, or EP is the polar of P 

with respect to V. 



• 2606. (Proposed by C. W. Meeeipield, F.R.S.)— The developable cir- 
cumscribing two surfaces of the second degree touches either of them along a 
curve, which is its intersection with another surface of the second degree. 



I. Solution by the Rev. R. TowKBEin), F.R.S. 

If U and V be the two quadrics, U' the polar reciprocal of U with respect 
to V, and V the polar reciprocal of V with respect to U ; then, for every 
plane tangent to both U and V, the point of contact with U being a point 
on V, and the point of contact with V being a point on U', therefore, &c» 
(See Salmon's Geometrif of Three Dimensions, 2nd Ed., Art. 207.) 

If aa^ + JiB« +073 + d82 ^ q and aV + b'e? + c V + d'h^ « be the equa* 
tions of U and V referred to their common self-reciprocal tetrahedron, 
then are 



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f!,.+ ^V+^7»+?«'-0 and li:«'+^/,.+ ''-y.+ |».-0 
abed a V c of 

those of U' and V referred to the same tetrahedron ; which, as is otherwise 
evident geometrically^ is consequently self-reciprocal with respect to the 
whole four surfaces at once. (See same. Art. 206.) 



II. Solution hy the Fboposeb. 

Referring the two surfaces to the centre of one of them, and to axes 
parallel to conjugate diameters in hoth, their equations may he written 
thus; — 

U - aiar,«+iB,yi» + 7i2i'' + «i = 0, 

The tangent plane of U is aia?iaf + /Biyiy + yiZiZ + Jj = 0. 
The tangent plane of V is 

03(a?2-0(a?-0 + /32(y2-«»)(y-»n)+'ya(«2-«)(2^-«) 

+ 8* - oa 2* - /32 m' — 72 »' - 0, 
or OsCar— Oay + ZSjCyj— iii)y + 'y,(«j— n)« + J2-a8te2-iB2«yj-7i2fiM!j = 0. 
And, if these are to he identical, 

«i^i ^ ^lyi ^ 71^1 ^ ?} 

ojCaj^-O 0iiy2-fn) 72(«^J-») «2-a2(^2-/32»»y2-72W 
Suhfliituting those values ofxj^yiZi in U, we get 

«2'gi(^2-0» ,. /32'8i(y2-«>)' ^ 72^81 (g2-«y 
«i i3i 7i 

+ (82-«2^2— /32«y2-72»»«2)' = 0. 

Hence the developable of two quadrics touches each along a curve through 
which another quadric can be drawn. 

Conversely, the envelope of the tangent planes to a quadric, along its in- 
tersection with another quadric, is the developable circumscribing the first 
and some other quadric. 



2804. (Proposed by S. Tebat.) — A strdght pole stands vertically on a 
slope inclined to the south. If it be broken at random by the wind blowing 
in a given direction, so that the upper end of the pole rests upon the slope, 
determine the probable area of the triangle thus formed ; and deduce the 
result for a horizontal plane. 

Solution hy the Pboposeb. 
Let a be the length of the pole, x the length of the portion standing, a the 
inclination of the slope to the vertical, j9 the azimuth of the wind measured 
on the slope, and ^ the angle between the standing piece and the base of the 
triangle. Then cos ^ » cos a cos i3, and 

A = ia?8in^{afcos^ + (a2— 2aa:+«'c0B'^)*}. 



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The gmtflst valae of « if f — , which is also a menBore of the number 

1 -|- BID^ 

of trianglee. Hence we find 

y ^"* 5 (8(l+8in4»)» 8co8?^ 2oo(i*^ 2coe»^^ "^i* 
an^ therefore the average area of the triangle is 

ia>8in * a + «n A) l_£?li 1 + ^ ''"'^ lojr cot 4*1. 

• '^^ ^M3a + 8in^)» 3oo8>4»^2co8<^ 2 cot* ^^ ^^3 

When a«iir, we have ^»^; and expanding logcot^ ■• ^log±^l5?^> 

1 — COB^ 

we find the average area « ^aK 



2618. (Proposed by the Bev. K. M. Fibsbbs, M.A.)— A pack of n dif- 
ferent cards is hdd, race downwards, on a table. A person names a certain 
card, that and all the cards above it are shown to him and removed ; he 
names another, and the process is repeated. Prove that the chance of hia 
naming the top card during the operation is 

liV H* E" 

Solution hjf Pbofbssob Whitwobth. 

Let P^ represent the chance of the top card being named in the series of 
operations on n cards. 

At the first operation any of the n cards ia equally likely to be named. 

If the top one U namedy what is required is done, and the ii posteriori 
chance of success is unity. 

j[jf the second be namedy there are «— 2 cards left, and the d posteriori 
chsince of success is P„.2* 

J^the third he named, then the d posteriori chance of success is Pn.a, end 
soon. 

If the last card be named, the chance is zero ; therefore 

or i»P„ = 1 + Pi+Pj+ .... +P„.3+ P,_2- 

Similarly (writing «— 1 for n) 

(»-l)Pn-i « 1 + Pi + Pa+ .... +P„.85 
therefore, by subtraction, nP^— (»— 1) P^_i - ^ii-2» 

»(Pn-Pn.l)=-(Pn-l-Pn-2)- 



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MaUaply both ndes by (—1)" |»-1 . then 

(-i)"lji(P»- P.-i) - (-1)""' t:l(P«-i- P»-8)- 

Now, writing n^l, »— 2, ii^3, &c. saooessively for n, we obtain, ex equdU, 
(-!)• l«.(P,-P,_i) - tl(P,-PO. 
Bnt it is obvions that Pi*l and Ps»i, therefore 

(-1)" WP„- P,_l) - -1. or Pn-P.-!' 



.(-1)-' 



Writing 2, 8, 4^ &c. saccesmvely for », we obtain 

P,-P.=._i-, P,-P,= + |., P,-P,»_l 4c 

P-P .- (-1)"-' 
Therefore, by addition, 

11 11 11 t 

With the notation of Questions 2637, 2648, this chance (of naming a top 
card) is l^e^^, and the chance of not naming a top card is e'^ • 



2813. (Proposed by J. J. Waleeb, M.A.)— The equation of a conic 
referred to an axis and tangent at vertex being ax^ + bi/^ + 2dje * 0, if the 
conic be turned about the vertex in its own plane through a right angle, the 
locus of the intersection of any tangent in the original position with the same 
line in the new position of the conic is the (bicircular) quartic 

and the corresponding locus for the normal is the sextic 

What does this latter equation become in the case of the panibohi ? 

I. Quaiermon Solution by W. H. Latxbtt, B.A. 
1. Taking for simplicity the equation to the conic in the form 
6V + ay-2a^«, 
the vector equations to the tangents are 

p «■ a + ocosT+i3 8inT + jr(— asinr + iSoosT), 

p » ?jS(1+cost) aunr+y (^i3sinr+ -a cost]; 

D a \o a J 



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72 

and that these may coincide, we mnst have 

a? (a^ sin' t + ^ cos' r) — a (a sin t + 6 cos t) + (a'— 6^) sin r cos r + a& ; 

whence substituting, and putting o'sin'T + i'cos^T «= P, we have for the 
required locus 

f».P Bs [fta(6cosT— a8inT) + a/B(6cosT + a8inT)J (1 + cost). 

To transform this to a Cartesian equation, we have 

ar.P , . J v.P , 

■ ttdcosT— asmr, and iJ . = ftcosT + asinT ; 

ab (1 + cos t) ab (1 + cos t) 

therefore ^^y. ^^-'^il^co-rf (,j^ 

also (pB-yY P3 - 4a2&2 (i + cos t)« o' sinS r 

or (a?-i/)«P =2a«8in«T(a?» + i/2) (2), 

and (* + y) P — 2a6 (1 + cost) 6 cost; 

from(l), {«»+5^-2a(ar+y)}P- 2a2is[(l + cosT)3-2oo8T(l + co8T)} 

= 2a2ft2 8in«r; 

from (2). {*3+y«-2a(a:+y)} {x^ + y^ = j^^oWy^+j^ ^ 6»(a?-y)^. 

or J {a(a:8+y3) + 2rf(a?+y)} (x^+t/S) = d2(ar-y)2. 

2. The vector equations to the normals are 

p :^ a + a cos T + jS sin T +x(b'^a cost + a^fi sin t), 

f»'« -jS+ ?/BcosT asiuT +y(ha$ cost— aJasinr) ; 

ho a 

and that these may coincide, we have 

— aAa?(a*8in'T + 62co82T) = afdcosr— aMUT] +a5 + (6*— a')cosT sinT; 

whence substituting, and putting 

Q?ein^r + b^coB^T^V and a2(i + cosT)-62cosT = Q, 

we get for the required locus 

ahFp — sinT Q[5a(asinT + 6cosT)— /Ba(ico8T— asinT)^. 

To transform this to a Cartesian equation, we have 

a?P — sin tQ (a sin t + 6 cos t), yP « sin tQ (a sin t— 6 cos t) j 
whence we easily find that 

^ 4a2&a8ln2r<y P* 46in-*rQg(ag-yyco68T , ,_ ^^ 

* P« aS62l6sin«TQ^cos«T pa «« V» r» 

or b{a{x+ify+b(x^yy] {a(x^+i/^ + dix+y)y^ ia--by€P(xi~ff^. 

3. In the parabola our equations become 

Pa? = <Q[a<-5} and Py = <Q[a« + j}, 
where P and Q now equal respectively aH^ + 6^ and ^ aH^ + 6*. 



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Let r^ a 5 ^ at ^alf^ (where I is the latuB rectum); then we have easily 
a?R^=:T(2T-0S and yR/ = t (2t + Z) 8, 
where K^4fr^ + P, and S = 2t3 + ^; 

therefore Z ^±1!? =. §, and -2T-:/i±i!; 

a?+y 2 ar— y 

therefore 4 ^1±^' =. i (^-hy)H2(^~y)2 



II. Solution hy the Proposes ; R. Tuoeeb, M. A. ; and others, 

1. The tangent heing {aaf + (Q a? + hy'y -{■da/=0, the equation to the same 
line, when the conic is turned through a right angle, will be 

iy'a? — (aa/ + d)y-da/^ 0. 
Adding these equations, we get (aO'+d) (x—y) = — 6/ (a? +y) ; 
and eliminating y by means of the equation to the conic, 

(aa?' + d)2(*-y)2 + 6(aF'3 + 2(^y)(*+y)2 = (1). 

Again, eliminating y' between the two equations above, we have 

(aay' + rf)(a;3+ya) + ^(a.+y)„0 (2). 

The elimination of a/ between (1) and (2) gives the first locus, which in the 
case of the parabola evidently reduces to a (circular) cubic 

2. The equations to the normal in the original and new position of the 
conic are, respectively, 

iy'a?-(aa?'+«?).y+y {(a— 6)a^ + rfJ = 0, 

and (flw?' + d)a? + ^y+y{(a— ft)a?'+cfj - 0. 

Pursuing steps similar to those in the case of the tangent, the required locus 
18 easily obtained. In the case of the parabola, developing, reducing, dividing 
by a, and finally making a^O, there results 

2(jP-y)2{ft(aj'+y3) + d(»+y)} + (?(a?+5f)8-0. 
If the given conic be a circle; the locus evidently becomes a circle also. 



2839. (Proposed by J. J. Waleeb, M.A.) — In a triangle the bisector of 
the base is equal to the less nde and also to one-half of the greater side; de- 
termine the three angles. 

Solution hy the Pbofobbb ; R. Tuokeb, M.A. ; and others. 

Let D be the middle point of the base BC, A the vertex of the triansle, 
and AC»AD»4AB. Then AC must bisect the angle between AD and BA 
produced (Euc. VI. A), and Z CAD « 180°- I BAG, 

TOL, zi. Z 



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therefore I BAC - 2C, therefore 3C » 180^ -B, 
whence Bin 3C ■> Ssin C — drin' C « nn B. 

Again, rin C : sin B « AB : AC » 2 : 1, or sin C « 2dn B, 
whence 6 sin C — 8 ain* C » sin C. 

IMsregarding the irrelevant solution CbO or 180^, we have sin' C « {^. 
But co8BACi«l-2sin3C«-^, and sin' B « i sin' C - ^3. 



2818. (Proposed hy the Rev. J. Wolstxitholice, M.A.)— Given a circle 
S and a straight line a not meeting S in real points; O, O' are the two 
point-circles to which, and S, a is the radical axis ; two conies are drawn 
osculating S in the same point P, and having one focus at O, O^ respectively; 
prove that the oorrespon(ting directrices coincide. 



Solution hy B. TucKSB, M.A. 



Join OP, OT, and draw OK, 
O'K, perpendiculars to them ;. then 
it is plain that these lines intersect 
in K on the circle through O'OP, 
cutting S orthogonally; and K is 
the intersection of the directrices. 
From the centre S draw SL, SL' 
perpendiculars on OP, O'P, and 
liM, L'M' perpendiculars on SP; 
then (Salmon's Conies, Art. 243) 
OM, O'M' are the directions of the 
axes of the conies, and 




therefore 



SM « SP sin» OPS. SM' - SP sin^ O'PS ; 
SM sin' OPS OPa SO. 
SM' ^ sin«0'PS *" O'pa " SO'' 



that is, OM is parallel to O'M', and the ^ectrioes (the perpendiculars from 
K on the axes) ooindde. 



2781* (Proposed by T. Cottbbill, M.A.)— Find the conic envelope of 
the radical axis of two circles, one of which is fixed, whilst the other passes 
through a fixed point. If the radii of the drcles are equal, find the foci and 
axes of the envelope. 

Solution by the Rev. J. Wolstenholmx, M.A. 

Let the fixed point be origin, a^+^'— 2ax— 2fiy a the equation of the 
moving circle, and (j?— o)*+y' « r^ that of the fixed circle; then, if r' bo 
the given radius of the moving circle, the radical axis is 

A4? + fiy "OX + i (a^- f^ « 0, with the conation x' + /*' - 1^1 



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and for the envelope 



or the envelope is a conic having^ its focus at the fixed point, its directrix 
bisecting the tangents drawn from the fixed point to the fixed circle, and 
eccentricity the ratio of the distance between the fixed point and the centre 
of the fixed circle to the radius of the moving circle. Tf the radii be equal, 
the envelope is the reciprocal polar of the fixed circle with respect to a circle 
whose centre is the fixed point and radius {i(a-— r^) j *. 



2724. (Proposed by S. Txbat, B.A.)_Three smooth rings, P, Q, B, are 
thread on an endless string of given length, and constrained to move on three 
straight rods, OP, OQ, OR. To investigate the motion corresponding to a 
slight arbitrary disturbance of the system. 



Solution by the Pbofoseb. 
Let OP=fa?, OQ-y, OR = «, 

PQ = «, OR = V, RP « 10, 

81, 83, $8 *he inclinations of OP, OQ, OR to the vertical; 
and T the tension of the string ; 

POQ-a, 0PQ«4», 0PR=:4»', 
QOR-iB, OQR = x. OQP = x'. ^r 

R0P = 7, ORP^tf. ORQ-f. / 

Then, for the motions of P, Q, R, we have 

P (^ - ^ cos 5i^ + T(co8 ^ + cos ^') - . 

Q (^ - /^ cos 82 ) + T (cos X + cos x') « . 

R (^ _ ^cos 8a) + T(cos 4^ + cosf ) « . 
And eliminating T, 
P(co8,»-+co6f) (g _ jcos 8,)-B(coB4. + coi*') (^ -i? «»•».)- 

(4). 

QCoM^ + cos*-) (gf -jrco. 8,)-E(co.x+co6x') (^-i'ooe %) - 

(6). 

These are the general equations of motion, but their complete solution 
cannot be effected except for small oscillations. 

The geometrical relations give 

tfcos^ -fycosaa or, o cos x +2^006/8 "«y, looos^' +d?00B7 »«, 
foco8^'+«coe7«af, woosx'-f-^oosa ">y, « cos t{/ f y cos jS — iE. 




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Let a^b^c be the Tfilaes of w, ff, z, when there is equilibrium ; a', 5', & the 
▼alnes of «, v, w; and A, K\ /i, /Jtf, y, ^ thoie of cos ^, cos 4»', cos x> oos x'» 
cos ^, cos 4^. 

Assume dr — a + dr', y — ft+y*, « = c+2' (6), 

i« = a' + ii', ««6' + «', w = c' + fi/ (7), 

where j;', y', tf^ u*, v\ u/ are small quantities. 

Now «3=:a^+^-2a?ycoso, B3«y« + a8— 2y«cosi8, io»««? + aj*— 2jkfcos7; 
or, neglecting small quantities of orders higher than the first, 
aV= (a^bcoBa)a/+ (5-acosa)y, JV« (6-0008^8)^+ (c-^cos^jz', 

c'lc/— (c— 00087)/+ (a— c cos 7)0?'. 
But a~ft cos a » a', &c., 

therefore vf^Kx'-^fiy, «/-/ty + //, icZ-yz'+AV (8). 

Hence, from (7), 

« = c^ + KaZ+iiyf^v « 6'+/i^ + k'z', 10 « c' + xV + f*'. 
Since «'+i/+it/«0, equations (8) give 

therefore (.+xog + (M^/)f + (v^.^)^ - 0. 

Also,from(6), ^«^. ^-^, ^==f?!. 

Substituting these in (4) and (5), neglecting small quantities of orders lugher 
than the first, and putting 

A = ^(..04(V/-co.,)_(^4)a.,,')^'. 

D - {^(x+V)+^(XK-coBy)j^-i; (X+A')-J-(^V-coe«). 

B' - {^ Oi + MO +|- 0»v-«»'8)} ii±A'_^0' + ^')4 ^'^""^")' 

D ={f 0.-M')4&.V-cos.)}^'_(1.4)(l.^a 
we find, after reduction, 

v + v' ' dt* v + Z * c?^ 

==5rCPAco8 8i + EBcosJs) +^(PCcos8i+ RDcos^), 



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77 

= ^(QA'coB 83 + RB'cos 8j) + ^(QC'cos 8j + RIV C08 J,). 
Ag^n, putting 
\ - QR(A + \0* + RPCu + fi7 + PQ(i^ + i/)9, 

IC = {Q(F + |/)2 + R(fl + /)2](PAC08jj + RBC08^ 

-R(A. + A')(/i + /)(QA'co8 83+RB'cos5j), 
Z - {R0i + A*T + Q(i' + i'')2}(PCcos8i + RDcos«,) 

-R(\ + x0^f* + fi')(QCco8Ja + Rl>'co8«a), 
*'- {R(X + V)a + P(i' + i/)2}(QA'co882+RB'co8 8s) 

-R(x + V)G^ + /)(PAcos8i + RBcos5a), 
r= {P(i^ + k')» + R(A. + V)'}(QC cos Jj + Riy 0088a) 

-R(A. + A')0* + /*')(PCoos8i + RDcob58)» 
wefinaUyget *(v + v)^-^(Ara' + Zy} « (9), 

»(''+»o^-j(*'*'+«y)-o (10). 

Hence, for small oscillations, the motion of the system is given by the 
simultaneoos equations (9), (10), and will consist of two independent vi- 
brations, of which the periods are 

where jf, rf' are the roots of the equation Vi? + (*— 2') 1 — '• 



2642. (Proposed by C. W. MBfiBiFiBLD, P.R.S.)— If three surfaces of 
the second degree meet in four points, at each of which their three curves of 
intersection have a common tangent, these four points lie in one plane. 



Solution "by the Pbofobeb. 

Write the equation of the quadric in its general form 

P = ax^+hy^-¥c^-¥d^+2lyz-¥2mzx-¥2nxy + 2px-\'2qy-¥Zrtwm0t 

and use suffixes to distinguish the three sur&ces Fi, Fj, Fs* 

Taking our origin at one of the points, and making the axes of ao and y 
pass through two others (2A, 0, 0) and (0, 2Jc, 0), we get {2»0 and j9a ^afi, 
q^^lk, where h and Ic are independent of the suffix. Take, moreover, the 
common tangent at the origin for the axis of x, and we have r^ r^ fs eacU 
=0. 



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The tangent planes at the origin may therefore be written at 
aihx + hiky ^0, o-Ao? + bjity ■■ 0, a^ + bJhy^O; 
and as these are to have a common intersection, we get 

«8 " ^ ' 

Treating in a similar manner the points (2h, 0, 0) and (0* 2k j 0)y for 
which the tangent planes are 

(or— 2/i) ah -{-y {2nh— hk) + z, 2mh »> 
and a (2nk- ah) + {y-2k)hk +z. 2lk - 0, 

we get the tother relations 

fli-g« ^ wi-wg ^ 2(ni--iia)A-(&i-b2)lr 

Os f»8 2njA— 5j* 

ii — Jji l\—h _ 2(»i— > ig)^— (fli — flg)A 
^3 " /s " 2jij*~a,A ' 

and since flZlf? - hzh, these are all equal, and so is also ^IZ!!!. We 

thus get the following reductions in the general equation :— 

d = 0, r - 0, £» . £2 = ?? « -A, £1 « i» - i» « -*, 
ai og aa ai Oj Og 

and we may therefore write the three quadrics as 

Pi « aypi^ + huf^ + ci«2 -|- 2iii/« + 2mi«a? + 2n^xy-~ 2aihx^2ibiJcy — 0, 
Fj - aja?3+ 

p, = f!ii:f?x2+ 5i-:^y«+c3«2+ 2^1:1^ yz+ -0, 

99 9 

the only outlying term being cg. 
But Fa- !izZ? « gives ^fiZfl+^aS « q, 

or there is a qnadric through the intersection which degenerates into the 
double plane ««0. This proves the theorem, which is easily reciprocated. 
The investigation, moreover, shows that, if there be three such points, there 
will also be a fourth. 



2771. (From WoLSTENHOLias*8 Book of Mathematieal Problems.)^ 
Prove that (1) if a straight line of length a be divided at random in two 
points, the mean value of the sum of the squares on the three parts is ^a'; 
(2) if the line be divided at random into two parts, and the longer part again 
divided at random into two parts, the mean value of the sum of we squares 



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on the three parts is ff a' ; and (3) if it is an even chance that n times the 
sum of the squares on the parts in (1) is less than the square on the whole 

I. Solution hy Stbfhek Watsoit. 

(1.) Let jf, y, and a—x—y he the parts. Then the total of cases is a«, 
and doabling because of the interchange of x and y, the average is 

o 

(2.) Here the number of cases is 

and the average is 

O 

(3.) In this case, let a?, i(a— «)— y» \{a^x) be the parts; then 

aj2 + i(a-ar)2 + 2y«<£! (1); 

therefore y< +i(TOa2-3«2)*, 

2 2 

where m wm ~ and % s x—^a. Hence the number of positions y can 

n 8 

take for each value of x is (ma'-3c^^» and therefore the chance p of (1) 
being fulfilled is, since dx^dz^ 

therefore » -•- — ^—r^ (whenp=i)^ ■^^' 



2ir + 3/>V'3' "^ '^ ' 4ir + 3V'3 

II. Solution hy the Bey. J. Wolbtestholme, M.A. 
(1.) The answer to this pert; of the question is 

/** /*"{af»+(y-«)8 + (a-y)2} dxdy-^ T C^'dxdy 
• c 

- 2a» /'^{(l + arS)(l-a?)-(l + fl?)«(l-*3) + |a-«»)}cfaf 

-*'»^/'\2-6« + 9««-5«») - 20^(2-8 + 8-4) - ia«. 



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(2.) The antwer to Uiii part of the qnestaon is 





9 V 8 24 64/ 72^ '72 

(8.) If AB be the rod ; P, Q the points of divuion, . x — 3 ^ 

P nearer A; AP « x, AQ - Y; then let no find the ^ 

chance that af»+(y-a?)»+(a-.y)a < f!, ^ ^c 

where x>{S<ay y>x<a. 

Take axes Ox, Oy inclined at 120°; measure 

OA « OB « a, and complete the rhombus OACB 5 o^ b sS 

then to every possible division of the rod corresponds 

a point in the triangle OAC, and to every favourable divimon a point within 

the circle a'+y'— ay-ay+ ^ = ^; 

2 2m 

or, transferring to the centre X«-XY + Y»- |! - —. 

When n^2, this is the inscribed drde ; and when » >2, the circle lies alto- 
gether within the triangle^ and 

its area : area of inscribed drde » : ~ « 2 (3— ») : n. 

Mod 

mnd area of inscribed drde : area of triangle » «• : 3 y^S; 

therefore chance we are seeking is — ^ ^ ^ ^ if » » -.; 

^ 3nv^3 2 4ir + 3-/3 

and since this is greater than 2, it is the value required. If fi<2, the chance 
that n times sum of squares is less than the square on whole is 

|z±(3.«2.+ 2,-6.). where «n.-(|z^)». 

The chance that 3 times squares on parts > 2 square on whole is ^— j-t 

1 2ir 

„ 7 „ „ > 5 squareon whole is ^ + — -_. 



2465. (Proposed by the Rev. R. Townsbnd, F.R.S.)— If a, 6, c be the 
three sides of a spherical triangle, and In the radius of its polar drde, prove 
the formula 

tan* k ■■ (<^ ^ cos c sec g - 1) (cos c cos g sec 6—1) (cos g cos 6 sec 0—1) 
4 sin « sin («— 0) sin («— b) sin («— c) 



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SohUioH hy the FBOPOBBB. 

If |}^ g, r be the three perpendicalars 
of the triangle; ai and a^ hi and h^, 
01 and e^ the three pairs of segmenfaB 
into which tiiey divide the three sides; 
Ai and Ag, Bi and B^ Ci and C2 the 
three pairs into which they divide the 
three angles; pi and p^ qi and 93, r| 
and rs the three pairs into which they 
divide each other; and h as above; 
then, rinoe 

tan pi tan p3 — tan 9i tan gs » tan ri tan rs a — tan^ *, 
and sinee, by Napier's rule, 

tan />tan/>i«ftancitanc3»tan6tanooo8A 

tan Pi a sin a^ tan B^ « tan /> cot B cot C sec a, 
sin 6 nn c cos A cos B cos C 




therefore, at once, tan' h » -^ 



smBsinC* cosacosicoso' 



which, substituting for the fnnctions of the angles their familiar values in 
terms of the sides, gives immediately the above. 

A process exactly similar leads to the corresponding formula for a plane 
triangle; viz., 

sin B sin C 82« («— a) («-6) («-c) 

[See Townsend's Modem Oeometry, Vol. I., art. 168.] 



2877. (Proposed by Professor Stltxbtbb.)— If 

V ^yet+gtx+txy-k-wyg, Y '^fytt + ijtex+hixy + kafyg; 

prove that the resultant of u, «, XT> V is 



SohtHon ly the Fbopobxb. 

Suppose x+y^O, g^O, t^O; then all four equations are satisfied provided 
/-^«0. Hence (/-^) (/- A) (y-A) (/-*) (^- *) (k-k) is a fiwtor of tiie 
TCsnltant. 

Again, suppose x^y^^g'^—t; all four equations are satisfied provided 
f-^gwmh + k. Hence (/+^-A-*)(/+A-^-*)(y+A-/-*) is also a 
fitttor of the same. 

But furthermore^ it f-i-sf—h^k ^0, then at the pcnnt of intersection 
«»1, yai, i;»— 1, <«— 1, It may eaoly be verified that the planes u, XT 

TOL. XI. K 



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reipectiyely touch the cubic sorfiMses o, V . Either of these fiicte warrants us 
in concluding that some power of (/•i-g—h—k) Mffher tha» iheJtrH must 
enter into the resultant Uenoe 

(/-^)(/-A)(/-*)(^-*)G^-*)(A-*) 

is contained in the resultant. But the order of the resultant in these letters 
is evidently 8*4-8, t.e. 12. Hence the above quantity is the complete re- 
sultant. 

I obtained the theorem originally by ordinary Algebra^ I forget how, pro- 
bably by a direct application of the dialvtic process; but the preceding 
method is more instructive, as embodying m a simple instance the transcen- 
dental law familiar to eliminationists, that a»y Mii$f«2ar% qfrelation between 
loci inUrtecting in virtue of a condition implies the appearance of a power, 
higher than the first, of the characteristic of that condition in the complete 
resultant. The theorem was wanted in an enquiry connected with the sub- 
ject of rectifiable compound logarithmic waves. A vast extension of it will 
be given in the answer to a subsequent question. 



NOTB Oir TBI -LLTR JUDOB HaBOBBATB'B SOLUTION OT THB QlTIKTIO. 

By the Rev. T. P. Kiekman, M.A., F.R.S. 

One of the final equations at page 94 of the Judge's j^thumous treatise is 

- S(A<). 
Ifso,itispo8dbletofind ai •»• jSi « Ai, a, •»- iS^ » A3, &c., such that 

where P* ■* yi, and Q* ■■ yj. 

As ^1 + ^s » 2 (51), p. 86, is a rational and symmetrical function of the 2% 
P* + Ci^ is one also; that is, P* is a two-valued function, whose values are P^ 
and Q*. It follows that P is also a two-valued function of the a^a, whose two 
values are P and Q; for if P had more than these two values, PPPPP + 
QQQQQ could not be a rational and symmetrical function. Hence P -f Q ob 
2 ( AO is a symmetrical function of the «*s. That is, 

y| +y| - 2 (AQ = A2^ = A2 («-dO = 0, 

and yi«-y2» or ^i+ys « 2(51) « 0. 

This destroys the form of the conditioned quintic in g, at p. 82 ; for (51)i-0 
gives, as (21) and (81) are each » 0, (p. 78). (p. 7), 

(51) -46i«- 5*154 + ^6-0, 

or bg is expressible in bi and ^4. Thus it seems that whatHABaBBATB has 
been solving is not the general quintic 



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2870. (Proposed by the Rev. J. Wolstbnholkb, M. A.)— QWen three 
p<nnt8 A, B, C, and a oonio; two points P, Q are taken on the conic such that 
the pencil A f BPQC j is harmonic ; prove that the envelope of PQ is a conks 
toaching AB, AC at points on the polar of A with respect to the given 



Solution hy Abohxb Staklet. 

Since the intersections x and af of every connector PQ and the given lines 
AB, AC are harmonic comugates with respect to P and Q, the polar of one 
mnst pass through the other ; in other words, x and x^ are conjugate points 
relative to the conic. Conversely, every connector of a pair of conjugate 
points Xf x^ on AB and AC, being harmonically divided by its intersections 
P, Q with the conic, must be a tangent of the required envelope. But to a 
point a; on AB there is but one conjugate af on AC, and vioe versd; hence x 
and a/ are corresponding points of two homographic rows, and their connector 
xj/ envelopes a conic which touches AB, AC at the points which correspond 
to A, that is to say, at the points B, C, where AB and AC are intersected by 
the polar of A. This envelope has obviously, in common with the given conic, 
the four tangents at the points where AB and AC cut that conic. 

The following are amongst the most interesting special cases : 

1. If ABC be a self-conjugate triangle, the envelope degenerates to the 
point-pair (B, C). 

2. The envelope also degenerates to a point-pair (A, A')» if A be on the 
given conic, and A' be the intersection of the tangents at the points where 
AB and AC cut the conic again. 

8. If AB and AC touch the conic in B and C, the envelope will likewise 
touch it in these points. 

4. If A be at the centre of the given conic, the envelope will be a con- 
centric conic having AB, AC for its asymptotes. If AB, AC be themselves 
asymptotes of the given conic, then the envelope will be a concentric, similar, 
and similarly placed conic. 

Many well known theorems in conies immediately follow from the above 
special cases, wherein, it should be observed, AB and AC may be imaginary. 
For instance, if these lines pass through the circular points at infinity, then 
the envelope under consideration will become that of chords of the given 
conic which subtend a right angle at A, and this envelope will have A for 
its focus. If A be itself a focus of the given conic, then the envelope (by 3) 
will have in common with it this focus and the corresponding directrix. If A 
be on the conic, however, then (by 2) the chords which subtend a right angle 
at A will all pass through a point A' on the normal at A. (Salmon's Conies, 
Art. 181, Ex. 2.) 



2828. (Proposed by S. Tebat, B.A.)— A straight rod is divided into n 
parts in arithmetical progression, and equal partides are fixed at the points of 
division. If the system be made to vibrate about one extremity, determine 
the average length of the pendidum, neglecting the weight of the rod. 



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SolutUm by the Fboposxb. 

Let a be tbe length of the rod, L the length of the pendulnm for any 

adjustment of the particles ; «!» 14 ^n~l ^^^^ distances from the point 

of snspennoQ. Then 

i«i + «s + .... +«,_! " 36(«„)' 
Let Ui^ae, d^~«i ■> x-t-y, where y is the common difference. 
Therefore «„ «i»{2a?+(n-l)yj; 

+ ,^n(n-l)(2n-l)(3w;»-3»-l)5f«, 
3K) -- in(n-l)(2ar-y)+ An(ti-l)(2n-l)y. 
But lw{2af+(w-l)y} = a; therefore y « ^^?~^x » 
therefore 2(i«:) - g.g-^_|?^ «V+aii(ii-2)(ii+l)(3»+l)a; 

+ 2a2(n-2)(3»'-6ii + l)J, 
S(«,) - ^n(n+l)a: + ia(n-2). 

Now the number of pendulums is proportional to ^ this being the greatest 

value of X, Therefore the average length of the pendulum is 

a J S(«^ 10»(»«-.1)( 

n + l ^2(11-2))' 

If » be increased indefinitely, this becomes 

If the distances of the particles irom the point of suspension be in arith* 
metical progression, and one particle be attached to the lower extremity of 
the rod, the average length of the pendulum is 



^^j£-^{4(2»-r) + 2(» + l)log2}. 



2533* (Proposed by Professor Hibst.) — Find the envelope of a line 
upon which two given conies intercept segments which have a common 
middle point ; and find also the locus of this middle point. 



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I. SoluHon hy the Rev. J. WaumasnoTXi, M.A. 

If the two conies be Ufi+my^ + nz^ ^0^ Z'^-i-m'i/' + nV ^0, and 
vx+tHf+wz =0 be a gtraight line divided as required^ we must have 

(m»'-m'«) JL , + (nl'^ n'l) JL + (lm'~ I'm) JS- = 0, 
r— to fo— « «— « 

the tangential equation of the envelope. For the locos of the middle pointy 
we have '*The envelope of a chord of a conic bisected by a given straight line 
is a parabola of which that straight line is a tangent." Hence the points in 
which a ^ven straight line wUl meet the locus are the points in which it 
will be met by the two remaining common tangents to two parabolas ; two 
points : or the locus is a conic. 



II. Solution by the Veopobsr, 

Let (C) and (Cf) be the two given oonics, L one of the lines whose envelope 
is required, and I the common middle point of the two segments which 
(C) and (CP) determine upon L. It is obvious that the intersections of (C) 
and (CQ with L determine an involution of which lis a double point, the 
other being at infinity ; and it is well known that every conic passing through 
the intersections of (C) and (C) will cut L in a pair of cox^jugate points of 
this involution, that is to say, will intercept on L a s^meut which has the 
same point I for its middle point. One of the conies of this pencil, in fact, 
will touch L at I, another will have L for its asymptote, and consequently 
have its centre on L. Now the locus of the centres of conies of the pencil 
b^g aconic (2), there will be another conic which has likewise its centareon 
L, and that cenlare must obviously be the common middle point I. 

Conversely, every point on (2) is a point I on some line L. For, besides 
being the centre of a conic of the pencil, it is the middle point of one chord 
of (G). Two conies of the pencil, therefore, intercept segments on this chord 
which have a common middle point I; hence all do so, and the chord is a 
podtion of L. The required locus of I, therefore, is the conic of centres 2. 

Moreover, the above investigation has shown that the required envelope is 
that of the asymptotes of the conies of the pencil [(0), (C)}* Hence its 
class can be easily shown to be 3. For through an arbitrary point x at in- 
finity only one conic of the pencil passes, and therefore only one proper 
asymptote. The pencil however contains two parabolas, so that the line at 
infinity must be counted twice as an asymptote through x. 

The required envelope being of the third class, and having a double tan- 
gent at infinity must, of course, be of the fourth order. 



2835. (Proposed by T. Cottbbill, M.A.) — In a plane, if A, B are two 
p(nnts, and a point P describe a curve of the nth order, show that P', the 
mtersection of the perpendiculars in the triangle PAB, will describe a curve 
of the 2nth order with three multiple points of the order n. Explain why 
the curve corresponding to a circle through the points A and B includes its 
reflexion to the line AB. 



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SohOum hsf the Pboposbb. 

(L) JM AC, BC be the perpendicDlan to the lino 
AB at the points A, B. Then, if P be a pomt in the 
phme, not on the lines, and PA, PB be joined, the per- 
pendicnlars AX, BY on PB, PA will intersect in a 
point P' i the pair will conjugately correspond, since 
each point is derived ftom the other by the same con- 
struction. The pur lie on a line through C, which is 
made np of such pairs of points ; amongst the lines is the line at infimW. 

Let A, B, C be called principal points. Then, if P lie on the line AC, V 
coincides with A. Similarly, to a non-^ncipal pcnnt on BC the point B 
always correroonds. If P is a non-principal point on AB, the point C cor- 
responds. Thus to every non-principal point of the plane one and but one 
point corresponds, whilst to eadi of the principal points the points of a line 
or the line itself corresponds. To a curve, therefbre, of the »th order, which 
does not pass through a principal point, a proper curve of the 2»th order 
will correspond, having multiple points of the order n at the points A, B, C. 
The nature of the correspondence of such curves is more clearly seen by 
observing (2) that since the angles at X and Y are right angles, the points 
X and Y lie on the circle described on AB as a diameter, and that P, F* are 
harmonic conjugates to the circle. But the line PP' always passes through 
C, the pole t/t the diameter AB ; so that we have a particular case of quadric 
inversion, the general prindples of which have been fully expluned by Prof. 
HnuST in his Piper on Quadric Invernon contained in the Proeeedingt qfthe 
Boyal Society for March, 1865. Corresponding points are therefore inverse 
points. One peculiarity of this inversion is that a right hyperbola through 
A, B is its own inverse; and rince the pole of PP' to such a conic passing 
tluough PP' is on XY, it follows that tangents to a curve at P and its in- 
verse at V meet on XY. 

(8.) The algebraical equations of the correspondence are very simple. Let 
(«> v» y)i («'* v't ^) be the perpendiculars from PP' on the lines AC, BC, and 
AB, the Hue PF cutting AB in M. Then PM . F'M » AM . MB, so that we 
have the equations vf^v, t/s «, and y[y » uo. Also 



y^ y y \ yy 

The equation to the drcle AXB, the locus of coinddent inverse points, is 
im-y3 1« K - 0. The right hyperbola (axis AB) is uv+y^ «= H « 0. If 
(«, v,y) ==0 be the homogeneous equation to a curve, (uy, vy, uv) is its in- 
verse ; (t^, vj/, H) s is its own inverse ; the inverse of (icy, vt/, E) «* is 
its reflexion to the line AB, &c. 

In tradng a curve which has a rimple inverse, the (xy) system of coordi- 
nates u often convenient, in which x is the perpendicular from P on OC, O 
being the centre of AB, y remaining the same as before. If AB a 2a, the 
inverse coordinates are connected by the equations x*^x, y'y ^c^^o?. 
Thus the inverse of the drde (y + 5)' » a'»^, which touches the lines AC, 
BC, is (6y •»- o^'^ofif =. (a'— «*)y'» a quartic curve having cusps at A and B, 
and a double point at C. If the drde does not cut the line AB, the quartic, 
b^ its revolution round OC, wdl generate a surface in the shape of a bell 
with its mouth uppermost, containing a bowl which just fits it at the rim. 

The inverse of a circle through AB is its reflexion to the line AB, because 
any point and the reflexion of its inverse to the line AB are on a circle 
through AB, as is easily seen. 



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282L (From Wolstekholmb's Book of MathemaHeal J^roblenu.)-^ 
Show that the mean value of the distance from one of the foci of all points 
within a given prolate spheroid is ^a(3 + 0^, 2a being the axis and e the 
eccentricity. 

Solution hy R. TuOKEB, M.A. 

Let P be an element of the ellipse at a distance 
r from the focos (S), and the angle ASP ~ 9 ; 
then, by the reyolation of the carve about AS, 
the element at P generates a small ring equal to 
2iir sin 9 r A9 An The volume (V) generated by 
ASL « 4iro8(l-62)(i + tf)8(2-ff), andthevolnme ^^ 
(V) generated by A'SL « ^ira8(l-e»)(l-c)2(2 + tf); 

hence -^ ^ (2-e)(l^ey ..^ -ZL « (llfKlz^'. 

v+v ii ' vTv' 4. 

For the part to the right of the boun^ng plane through SL, putting 

ff » £lLll^, the average distance of P is 
l-.eco89 

^yy',*sm<..»<Jr-|(l±i)(8-8e+«») (,). 

o • o* 
The average distance for the poridon to the left is found by changing into 

-c; thus it equals % ("IZ-f \ (3 + 3e + c>) {B). 

2 \2+ €/ 

The average distance required must be determined proportionally to the 
volumes ; thus it equals 



2815. (Proposed by A. Mabtiv.) — ^A cask contains a gallons of wine. 
Through a hole in the top water or wine can be let in at the rate of 6 gallons 
per minute ; and through a pipe in the bottom, when open, the mixture can 
escape at the same rate. Suppose the discharge pipe is opened at the same 
instant that water is let in at the top, and t minutes afterwards the water is 
shut off and wine let in. Bequired the quantity of water in the cask at the 
end of ti minutes from the opening of the discharge pipe, and the length of 
time elapsed, both before and after wine was let in at the top, when the 
quantities of the two fluids in the cask were equal, supposing them to mingle 
perfectly. 

Soluiion by C* R: Rippik, M.A. ; R. J. Nslson, M.A. ; the Pbofobeb ; 

and others. 
Suppose both pipes to have been open for any time ti and let ^ be divided 
into 11 equal intervals, and conceive the alteration in the fluid to hare been 



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effiacM hj removing ■— gftllons in each intenral and replenishing it with — 
• n 

gaUona of water. 

Then, if V be the number of gaU. of wine at the beginning of an interval, 
and Y" be the number of gals, of wine at the end of the same interval, 

we have V"= V- — . — = V /"l - — "^ - Te (o being constant) ; 
an \ an/ 

wo that, if y be the quantity of wine at end of time t, then, nnca there 
were a gallons at first, we have 

v.«...(>-g)-. 

Hence, making n infinite, we have for the result in the proposed question, that 

the wine at end of time ^ » ac * , and therefore the water « a (1— c '). 
Denoting this by a\ we have, by the same process, that the water in the 

cask after a fhrther time ^ « o'c * » a'c ' ; and the wine and water 

-^ a a 
will be equal when at ^ ^ z ^^ ^ « r log 2 (1), 

and also when a'* ""^.i ®' ^'^h^'T (^)' 



2816. (Proposed by the Editob.) — Find three square numbers in arith- 
metical progression, such that the square root of each (a) increased or else {0) 
diminished by unity shall give three rational squares. 

I. SoluHon by Sakitiel Bills. 

If we put a-2pg + (2?*-g'), J « 2pg— rp^— ^2), and o-p'+j', then 
will a', 2^, 0^ be three square numbers in arithmetical progression. 

Now put 2!, ^, ^ for the roots of the three squares required in the 

y* y* y* 

question ; then in the part (a) of the question we shall have to find 
?? + 1, — + 1, — + 1, or CM? + y*, te+y", osr+y* all squares. 

y* y* y* 

Assume a»+^-(r+y)» (1), and 5d?+^-(«+^)' (2). 

From (1) and (2) we readily find 



X 



-^?^ -*y-^ 






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Substituting these values in cx+y^, it becomes 

as—br 4 (as — brY* 
which is required to be a square ; that is, we must have 

4cr» (r— *) (fis^br) t (br^—eu^ '^ a square. 
The above reduces to 

4V— 4Aci^s + (4ac + 4Ac — 2ab) r*«*— 4aer^ + a V « a square. 

Assume the above - J^r8-2cr»+ f^^-^+ 2c-a^ jS^. 

Squaring this expression and comparing it with the above, we have 
-iacr+aU - -4c (^^J^+ 2c-a) r + (?f£^+ 2c -<.)'.. 

By reducing this, we find r « ^"^ "^ J. 

We give the following numerical examples, remarking that p and gr must 
be so assumed as to make s positive. 

o 

1. Take p^2 and g»l: then a»7, b^l, <;a5; whence r ».#. 

2 

6 19 

Take #=2, then rs3; whence d? b — , and y «« . 

11 22 

This gives ?5i 1??2, 1?*2 for the roots of the required squares. 

2. As another example, take p^l and 9f'*>4 ; then we get a«89, &b23, 

c-665 which gives r«lZ». Take *=46, then r=47, a?- -?i, y « -5?^ 5. 
46 131 262 

whence we obtain for the three roots, }^^, 55215^, !*?§??»*. 

(5979)3 (5979)2* (5979)« 

3. For a third example, take p-^5, and q^2; this gives a » 41, 5 »< — 1, 
e»29; or taking b positive, as we are at liberty to do, we shall have d»41^ 

-|Q 

A*-l, c«29 ; whence we find r^ — *. If now we take *«2, we have r»]3s 

2 

whence « - ??5, y = A, ± » ?????. From these values we find the 
69 ^ 138* i/« 25 

rooU of the three squares to be 3157*44, 91565*76, 12945504. 

Another solution may be obtdned as follows : — 
Put a-2pg + (p2-gr«), 6«p2+28, c = 2pg -(;)»- g^) ,' 

then will a^, b\ c^ be three square numbers in arithmetical progression. Now 

let ff , —f £^ be the roots of the three squares required in the question ; 

y» yi ya 

then we have to determine x andy so that ax + tf'^, bx+y^, cx+t/' may all 
be squares. 

Assume oay+y* =* P, bx+y^ « m', ca?+y* = »'; 

then ?nf_!!fn?_:^Zy! -«,(-«< suppose); 

a be 

therefore P— y' — ffr«f, m^— y' « ftra/, i|3— y^ = cr*<. 

TOL. XI. L 



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AMome l-i-y-iar, I-y-t^ «i+y-*t, m-y-rt, n+y-ee, ti-y-r*. 
We matt now have S^ -■ at^gt -• hs^rl — c<— r*. 

wbenoe (<+A)j» + (a5-<»)*-c/(< + a) - 0, 

therefore (^- o^)' •«• 4e< (^ -f- a) (< + &) must be a square. This redaces to 

l« •!• 4c<* + 2 (2ac 4- 26e— o^) (> -I- 4a&c^ + a^i* ; 
whieh aamime ■■ (l> + 2of + 2ac + 2^— 2c*— oft)' ; 
then, by redocixig this, we find t^ \ (c— 6— a). 
By symmetry, we have * • J (6 — c — a), r *- J (a— 6 - c). 
Also^ y - 4(ar-*0 - i(a' + ^+c»-2a4-2ac-2ftc)i 

wbfinfiA f. . 8( a-hc-&)(a + ^"C)(a-5-c) 

wii«uoe ^^ (2aA + 2ac + 2Ac-a«-6«-c8)a ' 

Therefore the roots of the three required squares will be 

8a(a+c— 5)(g'|-6-c)(a— 6"c) 8/i(g-|-c— 6)(a + 6— c)(a— ft-c) 
(2a^ + 2ac + 2&c-o«-6«-c»)a ' (2a6 + 2ac + 26c-a«-6»-c»)» ' 
8c (g +c~ft) (g-l- &— c) (g ^ft— c) 
(2aJ + 2flk? + 26c— aa-6»-cS)« ' 

where a, 5, c must, to satisfy (a), be the roots of three square numbers in 
arithmetical progression, such that, taking them all positively, the greatest 
root must exceed the sum of the other two. 

The results g^ven in the former solution may be readily obtained from this. 

If the greatest root be less than the sum of the other two, the result 
would satisfy (^), that is, it would give three square numbers in arithme- 
tical progresnon, such that, if each of their square roots be taken from unity, 
the three remainders shall be rational squares. 

For example, take j>— 8 and g»2; then a»17, ft»13, o»7; and we find 
for the three roots 

ax _ 108224 bx ^ 78936 c« ^ 42604 
y* " (855)3 ' y* " (365)*' y* "^ "" (355)«' 

Other answers may be readily found. 



II. Solution by the Frofobeb. 

Let x,y,z be the roots of the three required square numbers, then these 
. have to satisfy the conditions 

aj»+«8-2y«, « + l-m«, y + l^n^, z + lm^p^ (1,2,8,4), 

where m,n,p are rational numbers. From (I), by the aid of (2), (3), (4), 

wehave 111? . 4^:1?, or ^^'^"'-Z ^ (it-Hp)(i>-y) ^^^ 

y + z x-y n2+p2«2 (»+«)(»»-») 

Assume the relations m + n ^r(n+p), m—nsss(n — /;) ; 
then from these we obtain 

2L_«_!L.«_P_('»i suppose") (6). 

r-» + 2r* r+» 2-r + * V * / 



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Substitttting these vfilues in (5)» it becomes 

;t5»(r-*+r»+l)« + (r*+l)2-H4i!Zi (T); 

and we have to determine r and s so that the second side of (7) may be a 
perfect square. In order to effect this, let r«— 1 «=/; then, patting for 

shortness' sake sf — ^ ff+ 2 + - j , (7) becomes 

h^^^r+sy-hi^ir-s+f) [=(r + *-25Fj8 suppose] (8). 

Prom (8) we find r = J(y-/) ; and thence, by the aid of (6), (2), (3), (4), 
we obtain the following general expressions for x, y, z, the roots of the re- 
quired squares : 

4(r< + r-y)(rg-g+gf) 4y(r + g-^) 40-r + g) (l + g-.y) 

where g = i/+l+ :?* ^ '^ H£f-f)> and * = —t/, 

/ being any rational number whatever. 

Taking /» 1, we obtun the numbers given as the results to satisfy the 
first conmtion (a) when the question was originally proposed, viz. 
. 264 1320 1848 
86l' 861 ' 861* 
Taking/=i, we find 

_ 42504 78936 _ 103224 , 

"* ~ (355)*' ^ " p5j^' "* " (365)3 ' 

hence these fractions are the roots of three square numbers in arithmetical 
progression which satisfy the second condition (jB) in the question, that is, they 
are such that, if the roots be taken from unity, the three remainders will be 
rational squares. 



2486. (Proposed by G. O. Haklow.)— Find a point on an ellipse such 
that the normal produced to meet the curve again may cut off a maximum 
area. 



I. Solution by Pbote&sob Whitwobth. 

Let PG be the required normal meeting the conic 
again in G. Let F be a point on the curve adjacent 
to P, and P'G' the normal chord at F. Let O be the 
point of intersection of these two normals. Then 
ultimately, when F' coincides with P, O is the centre 
of curvature at P. * 

Since PG is the normal which divides the area the "^c^^ 

most unequally, therefore the greater area cut off by 
PQ is a maximum. Hence ultimately area PP'O » area GG'O. But these 
areas are ultimately in the ratio of the rectangles OP . OF and OG . 0G^ 




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Therefore OP. OF* 00.00' ultimately. But if Cp.Cpf be semi-dia- 
metert parallel to OP and OP, we have 

OP.OG : Or.OG'- (y : Cp'*; 
therefore 0P« : OG^ - Cp» : Op", or OP : OG'- Cp : €/>•. 
But nltimatelj Co « Cp'; therefore ultimately OP » 00" or OP » OG. 
Henoe PG is the diameter of curvatare and the common chord of the ellipee 
and circle of curvature. Hence PG and the tangent at P make equal angles 
with the axis, viz., angles of 46^. Therefore the normal required is a normal 
inclined at an angle of 46® to the axis. 

OoBOUiBT. — ^The normal which cute off the smallest dosed area from 
any oonio b inclined at an angle of 46® to the axes. 

[A solution by Mr. Jimcurs has been given on p. 66 of Vol. X. of the 



II. SotutHmhyHeEDJTOig. 

Let C be the centre of the 
ellipee; PCQ the diameter, and 
PGH the normal chord through 
P; NPUand MHV the ordinates 
at P and H, meeting the circle 
circumscribing the ellipse in U 
and V; and UCW a diameter of 
the circle through U. 

Then it may be readily shown b\ 
that U, G, V are in the same 
straight line ; and by a well known 
property we have 

area AGP AGH APH 
' AGV "* AUV 

PHQ b 

■ — — ^ — f 
a 



areaAGU 



UVW 




a and b being the semi-axes of the ellipse. 
PHQ, and p for the angle VOW, we have 



Hence, putting S for the area of 



S - area PHQ . 1 (UVW) - ^oi (^ + sin ^). 



Now 



H«?, and 
PN b' 



CN 
GN 



b«' 



hence 



tan UGN » ^ tan 9, 
b 



and 



tanUCN = -tnn«; 
a 



therefore tan i^ » tan CUG » tan (UGN-UCN) : 



a2-6» 



2ab 



sin 2a. 



Hence S » ^oft (^ + sin ^), where 2a6 tan ^ = (a^— 6>) sin 29. 
When S is a maximum, we must have 

whence we find the angle PGN « ss 450. 



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Hence we see that the normal cbord PGH, which makes an angle of 45^ 
with the major axis, divides the ellipse into two segments, one of which 
(APH) is a minimum, and the other (BPH) a maximum. 



2527. (Proposed by J. Wilson.)— Find the locus of the intersection of 
two tangents to a circle, the chord of contact of which subtends a right angle 
at a fix^ point ; and also of the middle'^points of a system of chords of a 
circle, which subtend a right angle at a fixed point ; and thence show that 
the envelope of a system of chords of a conic, which subtend a right angle 
at the focus, is another conic of coincident focus ; and that the locus of the 
middle points of a system of chords of a conic which subtend a right angle 
at a fixed point is another conic. 

Solution by the Bev. J. Wolstbkholmb, M.A. 

Takinff the fixed point for origin, let equation of circle be (a?— c)"+^«a', 
and let & + my «« 1 be a chord. The lines joining the points of intersection 
to the origin have for their equation 

«* + y' — 2ca: (to + wiy) + (c" - a^) (te + my)* « J 
and if they be at right angles, 

2-2ci + (c«-a«) (P + m«) - 0. 
We may then write the equation of two parallel straight lines satisfying the 
condition in the form 

2(te + my)a-2c«(te+ii^) + (c»-a«)(P+ifi») - 0, 
and the envelope is therefore 
(2*>-2ca?+c»-a?»)(2i,2 + c«-a»)-(2ay-cy)«, or (j£=^ + -^ i . 

a conic whose foci are the origin and the centre of the circle. The middle 
point of the chord being the foot of the perpendicular from the centre, lies 
on the auxiliary circle of this ellipse, t. e. on the circle {2x — c)' + %' n 2a^— c*. 
The locus of the intersection of tangents at the extremities of such chords is 
the reciprocal polar of this conic with respect to a circle whose centre is a 
focus, and is therefore a circle. Of course the envelope of a system of chords 
of a conic subtending a right angle at any point is a conic with that point as 
focus; as we get at once by redprocating the property of the director circle 
of a conic 

If the equation of any conic be aa5'+6y'-*-'2A* + 2Ay + l « 0, and the 
straight Ime to + my » 1 subtend a right angle at the origin, 

a + h + 2(hl + km) + ^ + ni?-mi0, 
and the coordinates of the middle point are 

X ^ Y 1 IJi + mY 

kml-hm^ + 6l Mm-kP + am ^ bi^i-um^ " hl^ + cuH^ 

gmX-ftnr 
■(am» + b^)(lr/-A«/ 
or «X + i»iY«l, Z(lr + iY) = m(.i + aX)j 

therefore - = *^ « ^ ; 

h + aX k^bY aX^ + l\i + hX + kY* 

and the locus of the middle points is apparently a curve of the fourth degree. 



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2769. (PropoBed by A. Mabtin.)— From a point, taken at random within 
a triangle, perpendiculars are drawn on the sides ; find the probability that 
the triangle formed by joining the feet of these perpen^cnlars is acot^ 
angled. 

Solution hf the Rev. J. WOLSTXKHOLICB, M.A. 

In an acnte- angled triangle ABC, if O be the centre of the drcmnscribed 
drdes, and arcs of circles be described on each side touching the radii from O 
to the ends of the side, the points mnst lie without each of the arcs. The 
result does not seem to be very neat, and it becomes more complicated for an 
obtuse-angled triangle. 



(Proposed by W. S. Bttbnsidb, M.A.) — GKve a geometrical 
interpretation for the relation between the invariauta of two conies, viz., 

SolmUon by the Rev. J. Wolbtbnholmi, M.A. 

The interpretation is given in Salmon's Conie Sections, It is the condition 
that the 8 points of contact of the common tangents to the two conies may 
lie on two straight lines ; in which ease also the 8 tangents to the two conies 
at their common points will pass through two points, four (two from each 
conic) through one point and four through the other. An example is any 
ellipse and a concentric rectangular hyperbola passing through the foci of 
the ellipse ; but the common tangents are unreal. Another example of two 
circles of radii 1, 7, with distances between centres 10, has the common 
tangents real, but the common points impossible. 



2109. (Proposed by Capt. Claskx, R.E., F.R.S.)— Two lines are drawn 
at random across a convex closed curve; determine the chance of their inter- 
secting. 

Solution by the Pkopobbb. 

Let P be a point within the curve, PQ a fixed line. 
Let AB be any line crossing the curve, the normal to 
which, Pp, makes an angle 9 with PQ, and let the in- 
tercepted chord AB ^ u. Let a second line, whose 
normal makes an angle d', be drawn intersecting AB ; 
the number of such lines is +« rin (0— 6^ according as 
is greater or less than 6': the values of 0, ^ to range 
from to V. The number of lines then that intersect AB ia 

•t /**Bin(a-l^da'+ n /''sin(0'- «)<W= u(l-co8#) + i»(I + coBe), 




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or 2t(. If Fp^Pt the number of lines intersecting all linee paraUel to AB is 
2 rudp » 2A, 

where A is the area of the curve ; and consequently the total number of 
pairs of lines that intersect is 2irA. It remains to find the total number of 
lines that cross the curve. Let Pf p' he the perpendiculars from P upon a 
pair of parallel tangents to the curve, then p + p^ is the number of lines 
which cross parallel to those tangents, and the total number is therefore 



/■ 



(p+p')d9^V, 



where P is the perimeter of the curve. The required probability of inter- 
section is, therefore, l5^. 

pa 



2773. (Proposed by Matthew Collins, B.A.)— When a given angle 
rolls upon a fixed parabola, the locus of its vertex is well known to be a 
hyperbola having the same focus and directrix as the parabola. Conversely, 
when a given parabola rolls within a fixed angle, show that its focus and 
vertex describe lines of the fourth and sixth orders respectively; find the 
actual equations of both these curves, and thence show that the vertex of 
the fixed angle is a conjugate point on the former curve, and that the 
equation of the hitter, when the given angle is a right angle, becomes 

the sides of the given angle being the axes, and a equal to the distance of 
the focus from the vertex. 

Solution bjf the Rev. J. Wolstbkholme, M.A. 
The equation of a parabola touching the axes of x, y, which are inclined at 
angle «, bemg f-\ + f- ) » h the coordinates of the focus are 

X Y hk 

* ■ A " h^ + k» + 2hkco8w 

and the equation of the directrix is 

jp (A + it; cos «) •!■ y (A -f A cos «) « Ait? cos «. 
Hence, if 4»% be the latus rectum, we have 

2a- ~ h^-*'k^ + 2hkcoBw " 

[ ( A + ifc cos w)' + ( A + A cos w)* — 2 (A + ifc cos w) (A + A cos w) cos w j * 



2A9A»BiM«« 



(AH^' + SAitcoswy 



y 



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whence the equation of the locus of the focus is 

a»(X» + Y»+2XY coe«> - X«Y«8iD««, 
a corre of the fonrth degree having a conjugate circalar point at the origint 
and having the fonr asymptotes X ^ +aoosecw, Y "- +aoosec«. The 
vertex is most readily found from being the point of contact of a tangent 
parallel to the directrix, which gives at once, if (X, Y) be the vertex, 

(kY)^ ^ (AX)* hi a (A» + ife» + 2Aifecosft>)* . 

h + kcoBtt ^ k + hcoBt0 '^ J?TW+2hFcoB0 " Bin2«» 7ik 

three equations from which to eliminate h, k^ In the case where « b iv, 
we have 

or ««-X«r»(X« + Y« + 8a»). 



2774. (Proposed by J. J. Waleeb, M.A.) — A central conic 
(i^o^+a^^ » a^l^ is turned in its own plane about its centre through a 
right angle ; prove that the locus of the intersection of the normal at any 
point on the given conic with the same line in its new position is one of the 
two sextic curves 

(a:« + y«)« {«*(*±y)»±6»(a?+y)«} « («^+5»)a(«»-yV. 



Quaternion SohOion hy W. H. Laybbtt, B.A. 
The vector equations to the two normals are 

f» » a COST 4- ^dnT + jp(J'o cosr + a^^ sinr), 

and p'^ ^fiooar a sinr + |^(&ai8 cost— odasmr); 

o a 

and that these may coincide, we have 

—«. 0^(0*^1* T + b'oori'T) « a&+(^— a')sinT cost. 

Whence, if a< sin' r + b^ cos' t » P, the equation to the locus becomes 

ab.V.p =■ gin T. COST (a*— 6^ ^b.a(aanT + ftcosT) + a.^(aanT— &cobt)J« 

To transform this to a Cartesian equation, we have 

P. a? arinT.008T(a>— b^(asinT + 6c()ST), 
P.y = sinT.oosT(a'— &3}(asin T— bcosT); 

therefore P' {a«(a?-y)«+ b«(a?+y)»} «-4a«b«sin«T cos^t (a»-6^, 
(»« + ySjS pa « 4 sin* t cos* t (a«-d«)*, ' 
(ajs«y»)3p4 « i6a262 8in«T coe«T(a3-^*, 

whence the required equation. Similarly for the hyperbola. 



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2244. (Propoeed by the late W. Lba.)— Form 11 symbols into sets, 6 
symbols in a set, so that every combination of 4 symbols shall appear once 
in the sets. 

2712* (Proposed by the late W. Lea.) — Form 15 symbols into sets, 6 
symbols in each set, so that every combination of 4 synibols shall appear once 
in the sets. 

2755. (Proposed by the late W. LEA.)^Form 16 symbols into sets, 4 
symbols in each set, so that every triad in the symbols shall appear once in 
the sets. 



Solution by the Rev. T. P. Eisexak, M.A., F.R.S. 

The problem of 4-plefc8 exbansting the triads of 16 is a case of the more 
^neral theorem which I gave in the Cambridge and Dublin Mathematical 
Journal, 1853, p. 42, thus : " 2** ypung ladies can all walk t(^ether in fourp 
day by day till every three have walked together." The proof is very simple. 
If n»2m, we take for our 4* symbols the repeating variations that can be 
made of them in 4-plets. .Then every triplet is completed into a 4.plet by 
the rule that no 4-plet shall have only three rtb places either like or unlike. 
Thus the triplet abbadd . bbcaab . cdaaaa (m=6) is completed into a 4-plet 
of axes by the six dddadc. When nf^Zm + l, we join to the 4*" symbols 
made with abed the 4"* made with abed, and add this rule, that no 4-plet 
shall have pnly three Italic or only three Roman symbols ; thus, if out of 
612 young ladies the three (abbd), (ahhb), and (adaa) choose to walk together 
on any day, their companion must be (adac), and I have shown that any 
triplet whatever will determine the entire arrangement for the day. 

To form 5-plets with eleven elements so as once to exhaust the 4-plets 
(Question 2244), we first form the 12 5-plets that contain the two elements 
Oa; thus 

Oal23 Oal47 Oal59 0al68 
0(7456 Oa258 0a267 0a249 (Oa) 
Oa789 0a369 0a348 0a357 
which are obtained by properly reading any one of the four squares of triplets 
in figures. We call this set of 12 (Oa). There must be in the solution a set 
of 12 (01), another (02), &c., another (12), &c., such that, if we form them 
all, we shall construct all the 11.6 quintuplets 10 times. Denote by 
(al, 4678) (Oa) the operation on (Oa) with the two circles al, 4678, that is, 
the exchange of a and 1, and the putting 6 for 4, 7 for 6, 8 for 7, and 4 for 8. 
Then we find that 

(01) = (al, 4678) (Oa), (al) - (01, 8764) (Oa), 

(02) « (a2, 5489) (Oa), (a2) = (02. 9846) (Oa), 

(03) -= (a3, 6697) (Oa), (a3^ = (03, 7956) (Oa), 

(07) = (a7, 1345) (Oa), (a4) « (04, 2197) rOa). 

(08) « (a8, 2166) (Oa), (a6) = (06, 3278) (Oa), 

(09) « (a9, 3264) (Oa), (a6) - (06, 1389) (Oa), 

(04) « (a4, 7912) (Oa), (a7) « (07, 6431) (Oa), 

(05) = (a5, 8723) (Oa), (a8) « (08, 6512) (Oa), 

(06) = (aJd, 9831) (Oa), (a9) « (09, 4623) (Oa). 

We have now every quintuplet containing and all containing a, in num. 
ber 12 + 18 + 18 = 48. 

If a - (4678), e-» = fl8 « (8764), and O^ « (47, 68) ; and the operation 



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(Oa, 47« 68) on (01) will give (al), and tfiee vend, (10) and (la) can be 
formed by readixig horizontally, yertically, and parallel to the dia^nals, th0 
two triplet squares in the right members following; 

(10) » 10 . a23 (la) » la . 023 
657 854 

849 679 

and we obtain la, 12, 18, &c from (10), thns 

(la) - (Oa. 47, 68) (10), (16) - (06, 29, 8a) (10), 

(12) » (02, 96, 54) (10), (15) « (05, 38, 42) (10), 

(13) - (03, 85, 79) (10), (IT) « (07, a4. 93) (la). 



B3,a6)aO), 
,7a,25)aO), 
,62, 37) (10). 

We hare now every quintuplet contuning 1. Next we writ^ by (2a) « 
(Oa, 94, 85) (20), (20) -20. lad (2a) » 2a. 103 
846 596 

795 748 

and we obtain, omitting (2a) and (21) ahready found, 

(23) = (08, 56, 47) (20), (27) - (07, 81, 84) 
(28) - (08, 17, 5a) (20), (29) - (09, 4a, 16) 

(24) = (04, a9, 73) (20), (25) » (05, 63, a8) 
(26) = (06, 35, 91) (20), 

This gives us every 5-plet containing 2. Next, we write (30) and (3a) » 
(Ofl,69.57)(30), 



(30) « 30.12a 
495 
687 
The process is continued thus : — 



(8a) -8a. 120 
467 
985 



- (06, 9a, 14) (80), 
(08. 61, 29) (30), 
(07, 42, a6) (30), 
which gives all containing 3. 
We now add the 54 following to the 12 above formed with Oa : — 
01657 02795 al854 a2748 12947 17958 
01849 02347 al679 a2397 12856 23875 
01254 02365 al257 a2368 12376 23496 
01379 02389 al349 a2345 12348 24765 
01358 03495 al356 a3467 12359 24589 
01629 08687 al829 a3958 18547 26789 
01278 04587 al246 a4759 13689 34789 
01346 04679 al378 a4986 14569 34568 
02846 05689 a2596 a5876 14867 35679 
If it is possible to form quintuplets with 15 letters so as to exhaust the 
quadruplets, there must be 26 containing the final letters de. These are 
easily formed by cyclically permuting the 18 elements 123 .... 90a6o under 
125 and 138. This gives us 

{de)^del26, d^2S6, .. ..c2el38, ({e249 

There must be a set (di) of 26, and another (el). We readily find 
(le) = (Id, 2358, 7a90) (de), 
(1<0 « (le, 8532, 09a7) (de), 
results so fiir analogous to the preceding. 



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Bat I cannot find the subjtitacions that will correctly give (2«), (2<i), &c. 
I shall be a^preeably Borprised to see this problem solved. It woatd be plea- 
sant to find it to be a simple evolution of the iamoiis fifteen yonng ladies 
whom I had the honour of introducing to the planet for the first time in the 
LadAfa and GentUmafCa Diary for 1850. To my original puzzle of seven 
arrangements of five triads so that every pair should once walk abreast. 
Professor Stlybstbb added a second, to make the school of 15 walk out 
every day in the quarter in 13 . 7 columns of triplets, till every three have 
once walked abreast. I gave, at the time, one solution of Ph)f. Stltbsteb'b 
question in the Cambridge and Dublin Mathematical Journal, It would be 
a memorable sight to see his 18 . 7 columns of 6 triplets made to perform the 
right-fiice, so as to present the 13 . 7 . 8 quintuplets here required. Any 
field-marshal could transform by a word the column of 5 . 3 into a dangerous 

Ehalanx of 3 .5; but I fear that a couple of generalissimos would find it a 
eavy strain on their tactics to turn out day by day for the quarter the 
exact quintuplets of Mr. Lba. 



2893. (Proposed by M. W. Cboittok, F.B.S.)— If there be {n) quantities 
a, 5, c, d, oc, each of which takes independently a given number of values 
aia^a^ .... bih^h^ . . . . &c. (the number may be different for each), if we put 

2(a) « a + 6 + c+rf + &c., 

and if for shortness we denote " the mean value of «" by M (;r), prove that 

M [2(a)] = M (a) + M(b) + M (c) + &c. « 2[M (a)], 

M C2(a)]2= {2 [M (a)]}'- 2 [M (a)]* + 2 [M (a«)]. 



Sohdion hy W. H. H. Hudson, M.A. 

Here a may have any of r values, ai^ oj, og ... ; b any of # values, 
hn &2, b^ ... ; e tjoj of i values, 0|, c^ e^ ... ; &o. &c. ; 
therefore a + 5-ho+ ... may have any of the rst... values 

ai+6j + ei+ ..., Og + Aj+Cj-f- ..., , 

whereof ai occurs in «^... sets, bi in rt„. sets, and so on; 

therefoi-e ^(^^^•^^)" (- M2a) - *<»-»(^-»-«a + »>0 + <>«-(^ + ^-*- >♦■)•^■.. 
rst„. ' rai,,, 

^ ai + c^+^ ^ Vijvt:^. ^ ^^^ ^ Ma+M6 + Mc + ... - 2Ma. 
r 9 

Agfdn, (2a)^ may have rst.., values, of which the type is 
...a? + 62+c»+... + 2aJ + 25o+... 
of will occur in H.,., b^ in rt,,,, and so on; ai^iwill occur in ^..., hiCi 
in r..., and so on 
Therefore M (2a)s « 

^...(ai» + Oa' + ...)+r<...(6i»+...) + 2<...(ai5i + ...) + 2r...(&iCi+...) 
ret 
^ ai« + Oa«+... ^ bf+bf+„. ^ ^^^ ^ Haih + aA+'") ^ 2(Vi+»-) . 
r • "** rs . 9t 



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100 



Now 



^ + -I- • . ..« 

r« st 

tberefm M(^)2 » M(aa) + M(&3)+ .... 

_ / fli + aa+... y_ / b, + b,+ ... y_ 
- 2Ma»+(5Ma)2-2(Ma)2. 



2887. (Proposed by Asohes Stanley.)— !• If a point P on a conic be 
eouuected with any two fixed points A and B in its plane, all chords which 
are divided hflrmonically by PA and VB will be concurrent. 

2. The locus of the point of concurrence when P is variable is another conic 
which has double contact with the given one on AB. 



Seiution hy F. D. Thomson, M.A. 

, 1. Let QB be the diord divided harmoni- 
cally by pa; PB. Then, constructing ns in 
the figure, 

-1 = p{3QTR} =:i p{cqdr} 

arR{CEDF}; 

therefore F is the pole of QR, and therefore 
QR passes through the pole of CD, i,e, 
through a fixed point. 

2. If O be the pole of CD, then when P 
"varies the locus of O is a conic. For consider 

the points in which the locus of O is cut by the tangent to the given conic 
through A. Let the tangent through A have L for its point of contact. Let 
BL meet the conic in M, AM meet the conic in N. Then the poles of LM 
and LN are the points on the required locus upon the tangent AL. Hence 
the locus is a conic. Also the tangents to the given conic at the points where 
AB meets it are seen to be tangents to the required locus. Hence the locus 
is a conic having double contact with the given conic at two points on AB. 
. The particular case when A and B are the circular points is given by 
Salmon, p. 257. 




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2728. <PropoKd by W. S. MoCat, B.A.)— Oiven Ibree planes and- 
tbdr poles with regard to a system of qnadrics, the locos of centre b a righ^ 
line. 

Solution hfi the Pboposbb. 

For the lines joining the vertices of a tetrahedron to the corresponding 
vertices 6f its polar tetrahedron belong to the same system of generators of 
an hyperboloid of one sheet. (Salmon's Geometry of Three Dimensiane, p. 
179.) Hence, the centre being the pole of the plane at infinity, we see that - 
if through each of the given poles we draw lines parallel to the intersection 
of the other two planes, the hyperboloid of which these are generators passes 
through the intersection of the three planes, and the lecas of centre is the 
generator of the same system at that, point. 



2597. (Proposed by W. H. H. Hudson, M.A.)— A right cone, whose 
weight may be neglected, is suspended from a point in its rim ; it contains 
as much fluid as it can : show that the whole pressure upon its surface is 
, ,. sin a cos 9 (cos (9 + a) /4 

where h and 2a are the height and vertical angle of the cone, and is deter- 
mined from 8 sin Si0 « 4 sin 2 {&— a). 



Solution hy Stephen Watson. 

Let C be the point of suspension; ABC a vertical 
section of the cone; BF a horizontal line cutting AC 
in D, and meeting a perpendicular from A in F; and 
O the middle of BD. Join aO, and take OG » ^OA ; 
then G is the centre of gravity of the fluid ; hence 
CO must cut OF perpendicularly in E, so that OE— 
^F. Put ZCBD — a, the area of the horizontal 
surface of the fluid «s S, and the area of the surface 
of the cone in contact with, the fluid » (A). 

Now BE » BC cos 9 «- 8A tan a cos a, 
BF -> BA cos (^— a-a) » A sec a sin (0 + a), 

and BO^-iBD-^^^^'*^^^'- *«^°« 




2sinBDC cos(0-a) 
hence, by substitution in BF-BO -4(BE-.B0) or BF + 3BO-4BB, 
we have sin (0 + a) cos (0-a) + 8 sin a cos a = 8 sin a cos ooS (0— a), 
.•. ^(8in20 + sin2a) + |sin2«-4sino[cos(20-o) + cosa} 

- 2 {sin 2a-sin 2(a-a)} +2sin 2a, 
therefore 3 nn 20 » 4 sin 2(0— a), the condition for finding 0. 
Again, if a, 6 (a « OB) be the semi-nxesi of the surface S, we have, by 



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102 



potlxDff |v--a-i-« for Ot md a for /S, in eqoafcioo (12), p. 895. VoL IL, of 
IHiviw emtioo of IT " ' " 



8 






Alao the distance of the centre of gravity of (A) from the horiioiital line 
BFia iAF-iBA8in(^*«-e)-iA8ec«cos(e+a) (1). 

Moreover, the orthogonal prqjedaons of the aoriaGef S and (A) upon the 
baw of the cone most hie equal, that ii^ 

(A)8in«-Soo6«, therefore (A)-^S; 

ana 
hence the whole pte w uie upon the oone% anrfoce ia 



210& (IVopoied by W. K. CuwrovD, B.A.) — Required analogoea in 
Solid Oeometry to the following propontioua in Plane Gkometry : — 

ia.^ The perpendicnlara of a triangle meet in a point. 
b,) The middle points of the dtagomtls of a qnadrilateral are in one straight 
line. 

(0.) The drdes whose diameters are the diagonals of a qnadrilateral have 
a common radical axis. 

(d,) Every rectangular hyperbola circumscribing a triangle passes throngh 
the intersection of perpendiculars. 

(0.) Every rectangular hyperbola to which a triangle is self-conjugate 
passes through the centres of the four touching drdes. 

(/.) mn (A + B) B nn A cos B + cos A sin B. 

(ff.) The sum of the angles of a triangle «■ two right angles. 

(A.) In any triangle ^ - 5L? - ?L2. 



SoUfUon hy the Pbofobib. 

I have an analogue for eadi of the four (cQ, (0), (f\ (jf), and more than 
one for each of the others obtained by extensions of Mr. Qsbbb*8 methods. 

(c.) A straight line cuts the faces of a tetrahedron ABCD in a, h, c,di 
the spheres whose diameters are Aa, Bb, Cc, Dd, have a common radical axis. 
Hence the middle points of these four lines are in one plane. 

Let ft conicoid whose asymptotic cone has three generating lines at right 
angles be called a rectangular conicoid. 

(d), (e). Every rectangular conicoid drcumscribing a tetrahedron whose 
perpenoiculars meet in a point, passes through the point. And every rect- 
angular conicoid to which a tetrahedron is self-conjugate, passes through the 
centres of the dght touching spheres. 



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103 

(/) oiitABC)-rin(ABD+BCD + CAD) 

» mn (BCD).C08 AJ) + mn (CAD). cos fib + sin (ABD) . cob ci>, 
where A, B, C, D are four tines in space, and 

on' ABC - 1, COB AB, 008 AC 
COS AB, 1,^ cos 1^ 
ooBAb, oosBC, 1, 
(j.) In the triangle case this should he written 
{BC)+(CA)+(AB)-0. 
The analogue is then ohvionsly 

(BCD)-(CDA) + (DAB)-(ABC) = 0, 
A, B, C, D heing any four planes. 
(A.) In any tetrahedron, 

AC.DB (dMfd a ._ Y^ ^ a ^ tt ^f! iv^n D, 

sinA&.sinbB V« cos A iabcd)^ 
where a, 5, c, rf are the feces, and cos^ A » 1, cos BC, cosBDI 

COS Sb, 1,_ cos CD 
cosBD, cos CD, 11 
(AB, &c denoting angles hetween planes). 



2661. (Proposed by W. S. Btoksidb, M.A.)— Determine the form of 
the solution of the differential equation 

2/(x)g + 8/'(x)^+{rW±«'}u-0 a). 

Solution hif the Frotobxr. 
The differential equation 

/(')g+*/'<'>|±-^-« <">• 

i. reduced to the farm g±»V-0 ^J "^""^ *^/:^^ ** *^* 
independent vwUble j whence, taking the lower rign, 

Agun, ffiibrentaating the equation (2) with regard to a>, 

ud writing U for ^ and 2/(») fiwr /(»), we tod the pven equation (1), 

the (ohiUon of which is therefore 



U-(A) 



ri^y-^Uhe"^"^]. 



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104 

8895. (Propoied by J. J. Wa£kxb, M.A.) — At any point on a cntped 
cnbic a tangent is drawn meeting the cubic in a second point ; from the first 
point a line is drawn touching the cnbic in a third point. Prove that the 
lines drawn from the cnsp to these points form with the cuspidal tangent a 
harmonic pencil. 



SoUaion hy W. H. H. Hudson, M Jk. 

Let the tangent at A meet the cubic in a, 

and from A draw AB tangent to the cubic. 

Let C be the cusp, CT the tangent thereat. 

It is required to prove that CB, Ca, CT, CA 

form an harmonic pencil. Take ABC as tri- 
angle of reference. The equation of the cubic 

18 y{la + mpif + fu?fi *= 0. 

[This may be obtained by taking the complete 

equation of 10 terms, and simplifying it from 

the considerations (1) that the cubic passes 

throneh A, B, C, (2) AB touches at B, (3) any 

line through C meets it in two points, then 

(4) the two lines that meet it in three points at C are coinddent.] 
la+mfi ^ is the equation of CT, 
Py + nfi^O is the equation of the tangent at A (Aa), 

and it meets the cubic where 2la + mjS » 0, which is therefore the equation 

of CA. From the form of these equations we see that CB, Ca, CT, CA form 

an harmonic pencil. 

[Mr. Thomson and the Pbofosbb remark that the theorem follows at 
once from Salmon's Higher Plane Curves, Art. 178.] 




2002. (Proposed by J. J. Waleeb, M.A.) — The circle aissing through 
three, points on a parabola, the normals at which co-intersecl]Mlways passes 
through the vertex ; and if the point of oo-intersection of the nomals describe 
a coaxial conic having the vertex as centre and axes in the ratio of m : n, the 
locus of the centre of the drde will be a conic having the focus as fMiitreb and 
axes in the ratio of 2m : ». 



Solution hy the Pbofobeb; W. Robbbts ; R. Tucebb, M.A. ; and other». 

Let (a, /3) be the common point of the normals, and y' — ^ax the para- 
bola ; then the equation to the drcle is readily shown to be 

aS+y»--(a+2a)a?~i/3y-.0 (1). 

From this equation wo see that the circle passes through the vertex; also 
that it U, f/) be the centre, 2*^ « o + 2a, 4^ = fi. Let n^a + m^0^ «= *» 
be the locus of (a, jS), then substituting 2(ir'— a) for a and 4/ for fi, 
» V* + 4m V^ » -^^ is the locus of (p/, yO^^ ^^''^c having its centre at 
{—0,0) and axes'in the ratio of 2m : ». 



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105 

2698. (Proposed by the Rev. J. WoLSTsyHOLiCE, M.A.)^A circle rolls 
with internal contact on a fixed circle of half its radius; prove that the en- 
velope of any chord of the rolling circle is a circle which reduces to a poiut 
for a diameter. 



Solution hy W. H. H. Hudson, M.A.; P. D. Thomson, M.A. ; 
R. TuOKEB, M.A.; and others. 

Let the diameter considered be the line 
joining the centres when the point of contact 
is A; when B is the point of contact let D be the 
centre of the large circle. Join DA; and pro- 
duce it to meet the large circle in E. 

Since BOA = 20DA, arc BA = arc BE ; 
therefore E is the point to which A has rolled, 
and DE is the new position of AC. Hence 
the diameter always passes through A. 

Now conceive a chord parallel to this dia- 
meter. It is always at a constant distance 
from it. Hence, if from A we draw a perpen- . . 

dicular upon it, this perpendicular is constant. It therefore envelopes a circle 
with centre A, which degenerates to the point A when the chord becomes a 
diameter. 




2875* (Proposed by the Rev. J. Wolstenholmb, M.A.)— Any tangent 
to a conic is^ of course, divided in involution by three other tangents, and 
the hues joining their points of intersection to one of the foci of the conic ; 
prove that the distance between the double points of the involution subtends 
a right angle at the focus of the conic. The locus of these double points, the 
three tangents being fixed, is a cubic having a double point at the focus, whose 
nature I have not examined. 



Solution hy Abches Stanley. 

Let A, B, C be fixed tangents to the given conic, an^X a viuriable one; 
moreover, let P and P' be the tangents from a given point p, (In the ques- 
tion p is a focus, P and P^ therefore pass through the circular points at iur 
finity, and all harmonic conjugates relative to them are orthogonal.) Now, 
by a well known theorem, the rays from p to the three pairs of opposite 
intersections of the quadrilateral ABCX are in involution, and to this invo- 
lution likewise belong all pairs of tangents from p to the several conic^ 
inscribed in the same quadrilateral. Now the given conic obviously belongs 
to this series, hence P, P' form a pair of conjugate rays of the involution; the 
double rays D^, Dj thereof must, consequently, be harmonic conjugates rela- 
tive to P and P', no matter what position X may have. This proves the first 
part of the theorem. ' 

With respect to the locus of the intersections of X with the double rays Dj 
and Dg, I observe that the latter are the tangents at p to the two conies 
inscribed in ABCX which pass through p. Hence, not only does X determine 
the harmonic conjugates Di, D2 relative to P and P', but, conversely, every 

vol. IX. N 



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106 

raeh line-pair D^ Bj determinee its oorresponding tangent X ; the latter is, 
in fact, the fonrtib tangent common to the given conic, and to that whidi is 
inscribed in ABC so as to toach Di (or D^) at p. It is easy, therefore, to 
determine the order of the locns, for through any point | of an arbitrary line 
L will pass one line-pair (Dj, Dj) corresponding to one tangent X, which cuts 
D say in w. Again, two tangents X pass through every point » ; and to 
them will correspond two line-pairs (Di, Dj) cutting L in four points |. 
Hence, between tne points ( and x there is a (1, 4) correspondence, and con- 
sequently there are five points in which x and ( are coincident. It is easy 
to see, however, that one of the latter is always on P and another on P': de- 
ducting these, which are not proper points of the required locus, we conclude 
that the latter is a cubic, of which p is a double point and P, P' the tangents 
thereat^ since on every line through p there is but one point of the locus 
whidi is not coinddent therewith, and on P (and P^ there are none, (When 
p is a focus of the given conic, it is also a conjugate point on the cubic.) It 
IS easy to find the nine intersections of the cubic with the fixed tangents A, 
B, C. Three of them will be the points a,b,c, where the sides A, B, G are 
intersected by the connector of p with the respective opposite vertices (BC), 
(CA), (AB) ; for instance, a is obviously a double point of the involation on 
the second tangent X which passes through it. The remaining two points of 
the cubic on A may be best found by conceiving X to coindde with it, or, in 
other words, to intersect it in its point of contact a ; for then a and a will 
be one pair of conjugate points of the involution on A, another pair will be 
the intersections (AB), (AC). The points where A cuts the cu bic, therefore, 
will be those which divide both the segments aa and (AB) (AC) harmonically. 
In this manner the order of the locus might have been more readily deter- 
mined. 



2003. (Proposed by W. H. Lavebtt.)— If A', B', C, IX, E', F be the 
minors which are the coefficients of A, B, C, D, E, F, respectively, in the 
different expansions of 
' A F E I 
F B D J and if ^1 denote either (A', B', C, D', E'. P0(\, a, y)*, or 
E D C| (A',B',C,D',E',F)(a,6.c)»; 

if ^ t= (1, 1, 1, —COS A, —cos B, —cos C) (A.. /«, i^); 

^-^ ^-($)(^)-(t)(^)' «<lQ'«'<lH«'-te»„>U., 

quantities involving y, \, and A, /», respectively; then show that, if a focus 
of the conic (A, B, C, D, E, F) (a, jS, 7)^ = be known to be in the line 
Aa+/iiS + iTr « 0, its coordinates {k\ /, •) are proportional to the minors of 

, and the equation to the diameter of the conic is 



the determinant \\ fi v 
IPQR 



a fi 



/d$i\ /d^\ /d^\ 
\daj \db ) \dc) 



= 0. 



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107 



SoluHon hy the Pbopobbb. 

It is eyident that if we find the pole of Xa+ ftjS + ^7 » 0, and firom this 
pole draw a perpendicalar to the Une, this perpendicular will meet the line in 
the focus. The coordinates, then, of the pole of the line are 

_^ !L i- (1). 

m it) ©) 

and the condition that aline 'La+Tdfi + 'Sy »0 •• (2) 

may he perpendicular to Aft+ fifi + try ss 0^ we should find to he 
L(X-/iCoe C-y cos B) * MOi- fcos A— Xcos C) + K(y-Xcoe B-/&COS A)-bO, 

H^h^'ith'^it)-' <»>• 

Also we have, from (1), mnoe {a' fif '/)Ib on the line^ 

'■ityitym-' '*>' 

therefore, from (2), (3), (4), we get (by eliminating L, M, N) for the equation 
of the perpendicular Va+Qfi + B,y ^ (5), 

and for the intersection of this with ha+zifi + ry^ we have (for the co- 
ordinates of the focus) 

\' ji' • 



Iqr| |rp| 

and since the coordinates of the centre are 



P Q 



.„_J 7 



(^\ (^\ (^\ 
\da) \db ) \do) 

we have, from (6) and (7), for the equation of the diameter 



.(6)S 



.•^ .P), 



V 



fi 

/ 



y 



fd^\ fd^\ /d^\ 
\da) \db) \do) 



= 0. 



2730. (Proposed by J» J. Wauebb, M.A)— If the determinant 

a e f a 

^ * J ^ VMUA. prwe that (erf) - ^^m^MB, ^here (6). (0, 

g % h d 

(A), ipd), ... are minors formed by omitting the rows and oo^umna in which 
the bracketed constituents stand. 



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108 



Solution by the Pbopobsb. 
In the Itaungtr of Mathematict, No. XIII., p. 27, I, have sbown that 

a*\); cong«;qnently, when D— 0, 



\m 



(bd) h 



be) 



^ ' (,bd)(bc)-ibky 

jei) {{e{)(bi)-{bJc)(dk)} ^■ (dh) [{d}i)(be)-{blc){ei)} 
~ (bd) lbe)-Q)kf 

.(WKjl+mffl («»p.26). 

(6) 

which is the type of six similar relations. 



(Proposed by the Rev. J. Womtehholme, M.A.) — Four fixed 
tangeuts are drHwn to a couic S ; three other conies are drawn osculating S 
in any the same point P, and each passing through the ends of a diagonal of 
the circumscribed quadrilateral ; prove that the tangents to these cf>nic8 at 
. the ends of the diagonal meet in one point P\ and that the locus of P' is a 
curve of the sixth degree and fourth class, having two cusps on every diagonal, 
and touching S at the points of contact of the four tangeuts. 




Solution hy F. D. Thomson, M.A. 

With the letters of the figure take the triangle ABC d 

formed by the diagonals as the triangle of reference, 
and let the equation to the line DEF he ax + by-k-cz-0. 
Then the coordinates of the points in the figure may be 
taken as 5^^^^"'^* 

D (0, c, —M, G (0, c, i), ^z:-'n2^^c 

E(-c, 0, a), H(c, 0, a), 

F(6, -a,0), K(5,«,0). \ \ 

By properly determining the constants, we can take as ^n,\ 

the equation to the given conic B 

ar»+y2+aj2«0 (i). 

Since DEF touches this, we get the condition 

a2+62 + c2-0 (ii). 

But, by Salmon's Conies^ Art. 251, the equation to a conic osculating (i) at 

the point (a^i^ V) is a^ + i^^ + z^+ {xaf +yy' + zz') {lx + my + nz)^0 (iii), 

where 2a?' + wy' + fi«'= (iv). 

If (iii) pass through the points D and G, we find, substituting the values of 
the coordinates, 

u.y(6i^c2) ^ gy ^ _ ^aV 



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109 

and therefore I "(g^-.V^ . 

therefore (iii) becomes 

Now, if (a?y«) be the pole of DG, t. e, of jp=:0, {xt/z)iB given by 

£? « 0, ^ - 0. 
dy dz 

These equations, after reductions by means of (ii) and a:^+y^ + 2'2 = 0, 

or eflx b^ ^ ^j 

which, from its symmetrical form, is evidently the same point as we should 
have obtained from either of the other two osculating conies which pass 
through EH and FK. Equations (v) give us 

x^ if'^ z^ ' 



(a^a?)* (6V)' (c^^)* (a»af)'+&c.' 
. therefore the locus of the point is the curve 

((Mr)* + (%)* + (c22)* = 0, or (a*xi + hy + C*z^f = 17a<bV«^V, 
a curve of the sixth order having two cusps on each diagonal, at the points 
where they are cut by the conic 

a^a^-i-by + c^z^ as 0. 
If we transform to tangential coordinates, the equation to the point given 

by(v)is ^K+^,i + ^y^O (vi); 

a* 0* c* 

therefore, to find the locus, differentiate, and we have. 

Z-Kda/^tliidy'^ L.ydz'^ 0, when af dx' -v if dtf ^ zfd^ ^ 0; 
a* 0* c* 

X, ^ af i/ ^ 

therefore — A.=s,^ttaa_-yj 

therefore, eliminating x\y\z\ ^ + — + ^ b (vii), 

A,» 11^ v^ 

the tangential equation to the locus. This is of the 4th class, and is satisfied 
by the lines ( +a, +&, -f c), and therefore touches the sides of the quadri- 
lateral. Also the pole 5 the line (x'^V) « XV (Mi/a + c*/»'*) + &c. « 0; 
therefore point of contact of (a, 6, c), t. e, of DEP, is aA + fe/* + ci' » 0, or in 

trilinears f^ + ^ « _, which is the point of contact of DEF with the given 

a b c 
conic ; therefore, kn. 

Cob.'- If the points F and E be the two real foci of the given conic, and 
E and H the circular points at infinity, B is the centre and DG the minor 
axis. Hence the osculating conic through EH and the point {afy'z') becomes 
the circle of curvature at afff9f\ that through DG and {p^y'v) an osculating 
conic through the imaginary fod; and that through FK and Qfjfi^ an oscu- 



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110 

latinff oonio ihroagh the real Ibd. Henoe the locos of the centre of cnrvatnre 
at (Jyi^) ia the same as that of the poles of the minor and major axes of the 
original conic with respect to the other two osculating conies. 



2787. (Proposed by C. W. Mebbifijexj), F.R.S.)<— The value of 
jfYTx'dxdyds taken over the whole volume of a tetrahedron throtkgh one 
of whose comers the plane of (yz) passes, is iV(ai' + 51^401*), where V is 
the volume, and ai, l^, ei are the distances of the middle points of the oppo- 
site edges from the plane of reference. 

SohOhn by the Pbofobxb. 

Let us first find the value of the integral taken over the volume of a 
pyramid having its ha§e in the plane of reference. Let the pyramid have a 
square base, and let one of its edges be perpendicular to its base. Let its 
height be h and the side of its base k. The Cartesian equation of one of its 

edges will be ^-j.^ k 1, and we shall have to find / sfif^dx subject to 

that equation. On substituting h(l—^ for y^ and performing the in* 

te^tions, we find its value to be -^ A^A' — -^ base x (altitude)^. It is 
evident that, whatever be the ^ape of the pyramid, we shall always have 

Jf/a^dxdydz » ^base x (altitude)*, 

provided the base of the pyramid be in the plane of reference. Moreover, 
the volume is \ base x (altitude). 

Now let ABCD be the tetrahedron, and 
DEFG the plane of reference. Produce the 
four planes of the pyramid to intersect with 
DEFG, as shown in the fig^ure, and let the 
distances of A, B, and C from the plane ot 
reference be a, jS, and y respectively. 

Since the pyramids ADFG and BDEF 
intersect so as to have the quadrangular 
pyramid DBCFG for a common frustum, we ^* 

ABCD - ADFG + CDEP- BDEG ; 
and also, if we call MI the integral Jfja^iadydn taken over the eorre* 
Bponding pyramid, we have 

MI . ABCD = MI . ADFG + MI . CDEF-MI . BDEG. 

But MI . ADFG « — DFG » ^ ADFG. 

30 10 

and similarly MI . CDEF . J^ CDEF, MI . BDEG = -^ BDEG. 

10 10 

Again, BDEG = ADFG + CDEF- ABCD; 

therefore XOMI.ABCD« (a«-i8«) ADFG + (r-/3^ CDEF + /8«. ABCD j 




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- Ill 

next, AFGD : ABCD = A AFG : AABC » AF . AG : AC. A6 

*=o.o: (a-7)(a-iB); 
also, CEFD : ABCD - aCEF : A ABC « FC . CE : AC . CB 

-7-7:(«-7)(^-7); 

. lOMI.ABCD a'(«^"ig^ . T^Ct'-ZB^) ^^ 
•• vol. ABCD (a-i8)(a-7) {y^a)(y-fi) ^ 

a— 7 0—7 a— 7 

-» o»+ 132 + 73 + /37 + 70+oiB 
-i(« + i8)« + i(/8 + 7)3 + i(7 + a)«; 
therefore MI . ABCD « ^V (ai« + V + ci«). 

MI is not a moment of inertia, being taken with reference to a plane instead 
of a line; hot the formnla, as it involves the squares, evidently holds for the 
moment of inertia aboat an axis. The common moment about the plane is 

iV(ai + 5i + C|). 
Moreover, the common moment and moment of inertia of a triangle about a 
line passing through one of its angular points are, severally, A being the area 
of the triangle, 

J A (flfi + bi + cO and ^A (a^ + b|' + cfl. 
The mmilarity of these forms is very remarkable. 



27(WL. (Proposed by the Rev. J. Wolstbkhoucb, M.A.)— In a right 
conoid whose axis is the axis of z, show that the radii of principal Gurva« 
ture at the point (r cos 9, r sin B, t) are given by the equation 

*'($)'*-»S!''*(»)T-!-(S)T- 



I. Solution by the Rev. J. L. Eitohin, M.A. 
The radii of principal curvature are given by the equation 

+ (l+/i«+3^-0 (1). 

The general equation of a conoid, axis of surface axis of ir, is z » ^ f^ ^ 

and since w^rco&9, y « r sin 9, this becomes z^^ (tan 9) »/(9) ; there- 
fore z is independent of r. Hence we find 

— n -. — Sin 9 — , — as -- cos 9 — ; 
dx r aO ay r dJ9 

andhence !+;.+,« .1+ J (g)^. J {^+ g)'}; also 

^*,-^Asirie^^-.^|^lsinel'^*:-lfisine^^^ 

da^ dx\r d9) dr \r d») dx d9 \r d») dx 

Effecting the differentiations and reducmg, remembering tha€ — is a func- 

d9 
tion of 9 only, we get 



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^z 8in»tf <P2 2 sin 2B dz ^ ^ , 



. ., , 6?z coR^ $ cPz 2sm2e dz . 
..m.lariy Tj^' -^ ^^ l^H'*' 

and ^ = _C2£^^ + !!!i|?^V hence 

'.'-'-M=tS*""S?.-"-»(S)'I 

4 fdz\^ 

"";iU; 

We find also that (1 +p2) «-2py*+ (1 + g») ri -. i ti. 
Hence (1) becomes 



_ 4 /d«y ,_ 1 d^ J \d9) i \ \dej ) 

?\de)^ f^^d^ r r* 

'^(S)-si-*(S)T-i-'(i)'r- 



= 0, 



II. Solution by the Pbofoses. 

Let the increments of r» 9, z, at a point Q near the ^ven point P, be 
A89, 80, Zz ; then the tangent plane at P is the limiting position of the plane 
containing the generating line Xsin 9 s= Ycos 9, Z^z, and passing through 
an adjacent point of the surface; its equation being taken Xsintf—YcosO 

+ Je (Z-«) = 0, ir is the Umit of (^ + ^)""^^^ or r -^ ^. But if p be the 

8z dO 

radius of curvature of the normal section containing the ultimate position of 
the line PQ, and p be the perpendicular from Q on the tangent plane at P, 

2p « limit of ^, 

V 
Now PQ2 « x'aea + ^28^ + 5^2, 

-(r + XW)^sinW+rf^M+f?f???+....^ 

^ ^ de \de de- z J 

^^ p r. — r.,z\h^ 



Co /dz\^ih V 

whence '^ " V "^ UJ » "^ 



dzy 
de) 



+ X2 



^ d0^ de 

and the principal radii of curvature are the maximum and minimum values 
of this for difi'erent vhIucs of A., and are therefore given by the-equation 

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